NCERT Solutions Class 12 Maths Chapter-7 (Integrals)Exercise 7.5

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-7 (Integrals)Exercise 7.5 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Exercise 7.5

Q1. Integrate the rational function : $\frac{x}{(x+1)(x+2)}$

Answer. $\begin{array}{c}\frac{x}{\text{Let}}\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\\ \Rightarrow x=A(x+2)+B(x+1)\\ \text{Equating the coefficients of}x\text{and constant term, we obtain}\end{array}$ $\begin{array}{c}A+B=1\\ 2A+B=0\\ \text{On solving, we obtain}\\ A=-1\text{and}B=2\end{array}$ $\begin{array}{c}\therefore \frac{x}{(x+1)(x+2)}=\frac{-1}{(x+1)}+\frac{2}{(x+2)}\\ \Rightarrow \int \frac{x}{(x+1)(x+2)}dx=\int \frac{-1}{(x+1)}+\frac{2}{(x+2)}dx\end{array}$ $\begin{array}{c}=-\log |x+1|+2\log |x+2|+C\\ =\log (x+2{)}^2-\log |x+1|+C\\ =\log \frac{(x+2{)}^2}{(x+1)}+C\end{array}$

Q2. Integrate the rational function : $\frac{1}{x^2-9}$

Answer. $\begin{array}{c}\frac{1}{(x+3)(x-3)}=\frac{A}{(x+3)}+\frac{B}{(x-3)}\\ 1=A(x-3)+B(x+3)\end{array}$ $\begin{array}{c}\text{Equating the coefficients of}x\text{and constant term, we obtain}\\ A+B=0\\ -3A+3B=1\end{array}$ $\begin{array}{c}\text{On solving, we obtain}\\ A=-\frac{1}{6}\text{and}B=\frac{1}{6}\\ \therefore \frac{1}{(x+3)(x-3)}=\frac{-1}{6(x+3)}+\frac{1}{6(x-3)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{1}{(x^2-9)}dx& =\int \frac{-1}{6(x+3)}+\frac{1}{6(x-3)})dx\\ & =-\frac{1}{6}\log |x+3|+\frac{1}{6}\log |x-3|+C\\ & =\frac{1}{6}\log \frac{|(x-3)}{(x+3)}|+C\end{array}$

Q3. Integrate the rational function : $\frac{3x-1}{(x-1)(x-2)(x-3)}$

Answer. $\begin{array}{c}\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\\ 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)\\ \text{Substituting}x=1,2,\text{and}3\text{respectively in equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{c}A=1,B=-5,\text{and}C=4\\ \therefore \frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{3x-1}{(x-1)(x-2)(x-3)}dx& =\int \{\frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)}\}dx\\ & =\log |x-1|-5\log |x-2|+4\log |x-3|+C\end{array}$

Q4. Integrate the rational function : $\frac{x}{(x-1)(x-2)(x-3)}$

Answer. $\begin{array}{c}\frac{x}{\mathrm{Let}}\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}\\ x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)...\left(1\right)\\ \text{Substituting}x=1,2,\text{and}3\text{respectively in equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{cc}A=\frac{1}{2},B=-2,& \text{and}C=\frac{3}{2}\\ \therefore \frac{x}{(x-1)(x-2)(x-3)}& =\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{x}{(x-1)(x-2)(x-3)}dx& =\int \{\frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)}dx\\ & =\frac{1}{2}\log |x-1|-2\log |x-2|+\frac{3}{2}\log |x-3|+C\end{array}$

Q5. Integrate the rational function : $\frac{2x}{x^2+3x+2}$

Answer. $\begin{array}{c}\frac{2x}{x^2+3x+2}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\\ 2x=A(x+2)+B(x+1)\\ \text{Substituting}x=-1\text{and}-2\text{in equation}\left(1\right),\text{we obtain}\\ A=-2\text{and}B=4\end{array}$ $\begin{array}{cc}\therefore \frac{2x}{(x+1)(x+2)}& =\frac{-2}{(x+1)}+\frac{4}{(x+2)}\\ \Rightarrow \int \frac{2x}{(x+1)(x+2)}dx& =\int \{\frac{4}{(x+2)}-\frac{2}{(x+1)}\}dx\\ & =4\log |x+2|-2\log |x+1|+C\end{array}$

Q6. Integrate the rational function : $\frac{1-x^2}{x(1-2x)}$

Answer. $\begin{array}{c}\text{It can be seen that the given integrand is not a proper fraction.}\\ \text{Therefore, on dividing}(1-x^2)\text{by}x(1-2x),\text{we obtain}\\ \frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{1}{2}\left(\frac{2-x}{x(1-2x)}\right)\end{array}$ $\begin{array}{c}\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{(1-2x)}\\ \Rightarrow (2-x)=A(1-2x)+Bx\\ \text{Substituting}x=0\text{and}\frac{1}{2}\text{in equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{c}A=2\text{and}B=3\\ \therefore \frac{2-x}{x(1-2x)}=\frac{2}{x}+\frac{3}{1-2x}\\ \text{Substituting in equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{c}\frac{1-x^2}{x(1-2x)}=\frac{1}{2}+\frac{1}{2}\{\frac{2}{x}+\frac{3}{(1-2x)}\}\\ \Rightarrow \int \frac{1-x^2}{x(1-2x)}dx=\int \{\frac{1}{2}+\frac{1}{2}(\frac{2}{x}+\frac{3}{1-2x})\}dx\end{array}$ $\begin{array}{c}=\frac{x}{2}+\log |x|+\frac{3}{2(-2)}\log |1-2x|+C\\ =\frac{x}{2}+\log |x|-\frac{3}{4}\log |1-2x|+C\end{array}$

Q7. Integrate the rational function : $\frac{x}{(x^2+1)(x-1)}$

Answer. $\begin{array}{c}\frac{x}{(x^2+1)(x-1)}=\frac{Ax+B}{(x^2+1)}+\frac{C}{(x-1)}\\ x=(Ax+B)(x-1)+C(x^2+1)\\ x=A{x}^2-Ax+Bx-B+C{x}^2+C\end{array}$ $\begin{array}{c}\text{Equating the coefficients of}{x}^2,x,\text{and constant term, we obtain}\\ A+C=0\\ -A+B=1\\ -B+C=0\end{array}$ $\begin{array}{c}\text{On solving these equations, we obtain}\\ A=-\frac{1}{2},B=\frac{1}{2},\text{and}C=\frac{1}{2}\\ \text{From equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{c}\therefore \frac{x}{(x^2+1)(x-1)}=\frac{(-\frac{1}{2}x+\frac{1}{2})}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}\\ \Rightarrow \int \frac{x}{(x^2+1)(x-1)}=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x-1}dx\end{array}$ $\begin{array}{c}=-\frac{1}{4}\int \frac{2x}{x^2+1}dx+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log |x-1|+C\\ \text{Consider}\int \frac{2x}{x^2+1}dx,\text{let}(x^2+1)=t\Rightarrow 2xdx=dt\\ \Rightarrow \int \frac{2x}{x^2+1}dx=\int \frac{dt}{t}=\log |t|=\log |x^2+1|\end{array}$ $\begin{array}{cc}\therefore {\int }_{(x^2+1)(x-1)}& =-\frac{1}{4}\log |x^2+1|+\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\log |x-1|+C\\ & =\frac{1}{2}\log |x-1|-\frac{1}{4}\log |x^2+1+\frac{1}{2}{\tan }^{-1}x|+C\end{array}$

Q8. Integrate the rational function : $\frac{x}{(x-1{)}^2(x+2)}$

Answer. $\begin{array}{c}\frac{x}{(x-1{)}^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{C}{(x+2)}\\ x=A(x-1)(x+2)+B(x+2)+C(x-1{)}^2\\ \text{Substituting}x=1,\text{we obtain}\end{array}$ $\begin{array}{c}\text{Substituting}x=1,\text{we obtain}\\ B=\frac{1}{3}\\ \text{Equating the coefficients of}{x}^2\text{and constant term, we obtain}\\ A+C=0\\ -2A+2B+C=0\end{array}$On solving, we obtain $\begin{array}{c}A=\frac{2}{9}\text{and}C=\frac{-2}{9}\\ \therefore \frac{x}{(x-1{)}^2(x+2)}=\frac{2}{9(x-1)}+\frac{1}{3(x-1{)}^2}-\frac{2}{9(x+2)}\end{array}$ $\Rightarrow \int \frac{x}{(x-1{)}^2(x+2)}dx=\frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1{)}^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx$ $\begin{array}{c}=\frac{2}{9}\log |x-1|+\frac{1}{3}\left(\frac{-1}{x-1}\right)-\frac{2}{9}\log |x+2|+C\\ =\frac{2}{9}\log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C\end{array}$

Q9. Integrate the rational function : $\frac{3x+5}{x^3-x^2-x+1}$

Answer. $\begin{array}{c}\frac{3x+5}{x^3-x^2-x+1}=\frac{3x+5}{(x-1{)}^2(x+1)}\\ \text{Let}\frac{3x+5}{(x-1{)}^2(x+1)}=\frac{A}{(x-1)}+\frac{B}{(x-1{)}^2}+\frac{C}{(x+1)}\end{array}$ $\begin{array}{c}3x+5=A(x-1)(x+1)+B(x+1)+C(x-1{)}^2\\ 3x+5=A(x^2-1)+B(x+1)+C(x^2+1-2x)\end{array}$ $\begin{array}{c}\text{Substituting}x=1\text{in equation}\left(1\right),\text{we obtain}\\ B=4\\ \text{Equating the coefficients of}{x}^2\text{and}x,\text{we obtain}\\ A+C=0\\ B-2C=3\\ \text{On solving, we obtain}\\ A=-\frac{1}{2}\text{and}C=\frac{1}{2}\end{array}$ $\begin{array}{c}\therefore \frac{3x+5}{(x-1{)}^2(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1{)}^2}+\frac{1}{2(x+1)}\\ \Rightarrow \int \frac{3x+5}{(x-1{)}^2(x+1)}dx=-\frac{1}{2}\int \frac{1}{x-1}dx+4\int \frac{1}{(x-1{)}^2}dx+\frac{1}{2}\int \frac{1}{(x+1)}dx\end{array}$ $\begin{array}{c}=-\frac{1}{2}\log |x-1|+4\left(\frac{-1}{x-1}\right)+\frac{1}{2}\log |x+1|+C\\ =\frac{1}{2}\log \left|\frac{x+1}{x-1}\right|-\frac{4}{(x-1)}+C\end{array}$

Q10. Integrate the rational function : $\frac{2x-3}{(x^2-1)(2x+3)}$

Answer. $\begin{array}{c}\frac{2x-3}{(x^2-1)(2x+3)}=\frac{2x-3}{(x+1)(x-1)(2x+3)}\\ Let\frac{2x-3}{(x+1)(x-1)(2x+3)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(2x+3)}\end{array}$ $\begin{array}{c}\Rightarrow (2x-3)=A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1)\\ \Rightarrow (2x-3)=A(2x^2+x-3)+B(2x^2+5x+3)+C(x^2-1)\\ \Rightarrow (2x-3)=(2A+2B+C)x^2+(A+5B)x+(-3A+3B-C)\\ \text{Equating the coefficients of}{x}^2\text{and}x,\text{we obtain}\\ B=-\frac{1}{10},A=\frac{5}{2},\text{and}C=-\frac{24}{5}\end{array}$ $\begin{array}{cc}\therefore \frac{2x-3}{(x+1)(x-1)(2x+3)}& =\frac{5}{2(x+1)}-\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}\\ \Rightarrow \int \frac{2x-3}{(x^2-1)(2x+3)}dx& =\frac{5}{2}\int \frac{1}{(x+1)}dx-\frac{1}{10}\int \frac{1}{x-1}dx-\frac{24}{5}\int \frac{1}{(2x+3)}dx\\ & =\frac{5}{2}\log |x+1|-\frac{1}{10}\log |x-1|-\frac{24}{5\times 2}\log |2x+3|\\ & =\frac{5}{2}\log |x+1|-\frac{1}{10}\log |x-1|-\frac{12}{5}\log |2x+3|+C\end{array}$

Q11. Integrate the rational function : $\frac{5x}{(x+1)(x^2-4)}$

Answer. $\begin{array}{c}\frac{5x}{(x+1)(x^2-4)}=\frac{5x}{(x+1)(x+2)(x-2)}\\ \frac{5x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}\\ 5x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)....\left(1\right)\end{array}$ $\begin{array}{c}\text{Substituting}x=-1,-2,\text{and}2\text{respectively in equation}\left(1\right),\text{we obtain}\\ A=\frac{5}{3},B=-\frac{5}{2},\text{and}C=\frac{5}{6}\\ \therefore \frac{5x}{(x+1)(x+2)(x-2)}=\frac{5}{3(x+1)}-\frac{5}{2(x+2)}+\frac{5}{6(x-2)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{5x}{(x+1)(x^2-4)}dx& =\frac{5}{3}\int \frac{1}{(x+1)}dx-\frac{5}{2}\int \frac{1}{(x+2)}dx+\frac{5}{6}\int \frac{1}{(x-2)}dx\\ & =\frac{5}{3}\log |x+1|-\frac{5}{2}\log |x+2|+\frac{5}{6}\log |x-2|+C\end{array}$

Q12. Integrate the rational function : $\frac{x^3+x+1}{x^2-1}$

Answer. $\begin{array}{c}\text{At can be seen that the given integrand is not a proper fraction.}\\ \text{Therefore, on dividing}(x^3+x+1)\text{by}{x}^2-1,\text{we obtain}\\ \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}\end{array}$ $\begin{array}{c}\frac{2x+1}{x^2-1}=\frac{A}{(x+1)}+\frac{B}{(x-1)}\\ 2x+1=A(x-1)+B(x+1)\end{array}$ $\begin{array}{c}\text{Substituting}x=1\text{and}-1\text{in equation}\left(1\right),\text{we obtain}\\ A=\frac{1}{2}\text{and}B=\frac{3}{2}\\ \therefore \frac{x^3+x+1}{x^2-1}=x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{x^3+x+1}{x^2-1}dx& =\int xdx+\frac{1}{2}\int \frac{1}{(x+1)}dx+\frac{3}{2}\int \frac{1}{(x-1)}dx\\ & =\frac{x^2}{2}+\frac{1}{2}\log |x+1|+\frac{3}{2}\log |x-1|+C\end{array}$

Q13. Integrate the rational function : $\frac{2}{(1-x)(1+x^2)}$

Answer. $\begin{array}{c}\text{Let}\frac{2}{(1-x)(1+x^2)}=\frac{A}{(1-x)}+\frac{Bx+C}{(1+x^2)}\\ 2=A(1+x^2)+(Bx+C)(1-x)\\ 2=A+A{x}^2+Bx-B{x}^2+C-Cx\end{array}$ $\begin{array}{c}\text{Equating the coefficient of}{x}^2,x,\text{and constant term, we obtain}\\ A-B=0\\ B-C=0\\ A+C=2\end{array}$ $\begin{array}{c}\text{On solving these equations, we obtain}\\ A=1,B=1,\text{and}C=1\\ \therefore \frac{2}{(1-x)(1+x^2)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}\\ \Rightarrow \int \frac{2}{(1-x)(1+x^2)}dx=\int \frac{1}{1-x}dx+\int \frac{x}{1+x^2}dx+\int \frac{1}{1+x^2}dx\end{array}$ $\begin{array}{c}=-\int \frac{1}{x-1}dx+\frac{1}{2}\int \frac{2x}{1+x^2}dx+\int \frac{1}{1+x^2}dx\\ =-\log |x-1|+\frac{1}{2}\log |1+x^2|+{\tan }^{-1}x+C\end{array}$

Q14. Integrate the rational function : $\frac{3x-1}{(x+2{)}^2}$

Answer. $\begin{array}{c}\text{Let}\frac{3x-1}{(x+2{)}^2}=\frac{A}{(x+2)}+\frac{B}{(x+2{)}^2}\\ \Rightarrow 3x-1=A(x+2)+B\\ \text{Equating the coefficient of}x\text{and constant term, we obtain}\\ A=3\\ 2A+B=-1\Rightarrow B=-7\end{array}$ $\begin{array}{cc}\therefore \frac{3x-1}{(x+2{)}^2}& =\frac{3}{(x+2)}-\frac{7}{(x+2{)}^2}\\ \Rightarrow \int \frac{3x-1}{(x+2{)}^2}dx& =3\int \frac{1}{(x+2)}dx-7\int \frac{x}{(x+2{)}^2}dx\\ & =3\log |x+2|-7\left(\frac{-1}{(x+2)}\right)+C\\ & =3\log |x+2|+\frac{7}{(x+2)}+C\end{array}$

Q15. Integrate the rational function : $\frac{1}{x^4-1}$

Answer. $\begin{array}{c}\frac{1}{(x^4-1)}=\frac{1}{(x^2-1)(x^2+1)}=\frac{1}{(x+1)(x-1)(1+x^2)}\\ \text{Let}\frac{1}{(x+1)(x-1)(1+x^2)}=\frac{A}{(x+1{)}^{+}}+\frac{B}{(x-1)}+\frac{Cx+D}{(x^2+1)}\\ 1=A(x-1)(x^2+1)+B(x+1)(x^2+1)+(Cx+D)(x^2-1)\end{array}$ $\begin{array}{c}1=A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+C{x}^3+D{x}^2-Cx-D\\ 1=(A+B+C)x^3+(-A+B+D)x^2+(A+B-C)x+(-A+B-D)\\ \text{Equating the coefficient of}{x}^3,x^2,x,\text{and constant term, we obtain}\end{array}$ $\begin{array}{c}A+B+C=0\\ -A+B+D=0\\ A+B-C=0\\ -A+B-D=1\\ \text{On solving these equations, we obtain}\\ A=-\frac{1}{4},B=\frac{1}{4},C=0,\text{and}D=-\frac{1}{2}\end{array}$ $\begin{array}{c}\therefore \frac{1}{x^4-1}=\frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}\\ \Rightarrow \int \frac{1}{x^4-1}dx=-\frac{1}{4}\log |x-1|+\frac{1}{4}\log |x-1|-\frac{1}{2}{\tan }^{-1}x+C\end{array}$ $=\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}x+C$

Q16. Integrate the rational function : $\frac{1}{x(x^n+1)}[\text{Hint: multiply numerator and denominator by}{x}^{n-1}\text{and put}{x}^n=t]$

Q17. Integrate the rational function : $\frac{\cos x}{(1-\sin x)(2-\sin x)}[\text{Hint}:\text{Put}\sin x=t]$

Answer. $\begin{array}{c}\frac{\cos x}{(1-\sin x)(2-\sin x)}\\ \text{Let}\sin x=t\Rightarrow \cos xdx=dt\\ \therefore \int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx=\int \frac{dt}{(1-t)(2-t)}\end{array}$ $\begin{array}{c}\text{Let}\frac{1}{(1-t)(2-t)}=\frac{A}{(1-t)}+\frac{B}{(2-t)}\\ 1=A(2-t)+B(1-t)\end{array}$ $\begin{array}{c}\text{Substituting}t=2\text{and then}t=1\text{in equation}\left(1\right),\text{we obtain}\\ A=1\text{and}B=-1\\ \therefore \frac{1}{(1-t)(2-t)}=\frac{1}{(1-t)}-\frac{1}{(2-t)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{\cos x}{(1-\sin x)(2-\sin x)}dx& =\int \{\frac{1}{1-t}-\frac{1}{(2-t)}\}dt\\ & =-\log |1-t|+\log |2-t|+C\end{array}$ $\begin{array}{c}=\log \left|\frac{2-t}{1-t}\right|+C\\ =\log \left|\frac{2-\sin x}{1-\sin x}\right|+C\end{array}$

Q18. Integrate the rational function : $\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}$

Answer. $\begin{array}{c}\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-\frac{(4x^2+10)}{(x^2+3)(x^2+4)}\\ \text{Let}\frac{4x^2+10}{(x^2+3)(x^2+4)}=\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}\end{array}$ $\begin{array}{cc}4x^2+10& =(Ax+B)(x^2+4)+(Cx+D)(x^2+3)\\ 4x^2+10& =A{x}^3+4Ax+B{x}^2+4B+C{x}^3+3Cx+D{x}^2+3D\\ 4x^2+10& =(A+C)x^3+(B+D)x^2+(4A+3C)x+(4B+3D)\end{array}$ $\begin{array}{c}\text{Equating the coefficients of}{x}^3,x^2,x,\text{and constant term, we obtain}\\ A+C=0\\ B+D=4\\ 4A+3C=0\\ 4B+3D=10\\ \text{On solving these equations, we obtain}\end{array}$ $\begin{array}{c}A=0,B=-2,C=0,\text{and}D=6\\ \therefore \frac{4x^2+10}{(x^2+3)(x^2+4)}=\frac{-2}{(x^2+3)}+\frac{6}{(x^2+4)}\end{array}$ $\begin{array}{c}\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1-(\frac{-2}{(x^2+3)}+\frac{6}{(x^2+4)})\\ \Rightarrow \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\int \{1+\frac{2}{(x^2+3)}-\frac{6}{(x^2+4)}\}dx\end{array}$ $\begin{array}{cc}& =\int \{1+\frac{2}{x^2+(\sqrt{3}{)}^2}-\frac{6}{x^2+2^2}\}\\ & =x+2\left(\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2}{\tan }^{-1}\frac{x}{2}\right)+C\\ & =x+\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x}{\sqrt{3}}-3{\tan }^{-1}\frac{x}{2}+C\end{array}$

Q19. Integrate the rational function : $\frac{2x}{(x^2+1)(x^2+3)}$

Answer. $\begin{array}{c}\frac{2x}{(x^2+1)(x^2+3)}\\ \text{Let}{x}^2=t\Rightarrow 2xdx=dt\\ \therefore \int \frac{2x}{(x^2+1)(x^2+3)}dx=\int \frac{dt}{(t+1)(t+3)}\dots \left(1\right)\end{array}$ $\begin{array}{c}\text{Let}\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}\\ 1=A(t+3)+B(t+1)\\ \text{substituting}t=-3\text{and}t=-1\text{in equation}\left(1\right),\text{we obtain}\\ A=\frac{1}{2}\text{and}B=-\frac{1}{2}\end{array}$ $\begin{array}{c}\therefore \frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\\ \Rightarrow \int \frac{2x}{(x^2+1)(x^2+3)}dx=\int \{\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\}dt\end{array}$ $\begin{array}{cc}& =\frac{1}{2}\log |(t+1)|-\frac{1}{2}\log |t+3|+C\\ & =\frac{1}{2}\log \left|\frac{t+1}{t+3}\right|+C\end{array}$ $=\frac{1}{2}\log \left|\frac{x^2+1}{x^2+3}\right|+C$

Q20. Integrate the rational function : $\frac{1}{x(x^4-1)}$

Answer. $\begin{array}{c}\frac{2x}{(x^2+1)(x^2+3)}\\ \text{Let}{x}^2=t\Rightarrow 2xdx=dt\\ \therefore \int \frac{2x}{(x^2+1)(x^2+3)}dx=\int \frac{dt}{(t+1)(t+3)}\end{array}$ $\begin{array}{c}\text{Let}\frac{1}{(t+1)(t+3)}=\frac{A}{(t+1)}+\frac{B}{(t+3)}\\ 1=A(t+3)+B(t+1)\\ \text{Substituting}t=-3\text{and}t=-1\text{in equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{c}A=\frac{1}{2}\text{and}B=-\frac{1}{2}\\ \therefore \frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{2x}{(x^2+1)(x^2+3)}dx& =\int \{\frac{1}{2(t+1)}-\frac{1}{2(t+3)}\}dt\\ & =\frac{1}{2}\log |(t+1)|-\frac{1}{2}\log |t+3|+C\\ & =\frac{1}{2}\log \left|\frac{t+1}{t+3}\right|+C\\ & =\frac{1}{2}\log \left|\frac{x^2+1}{x^2+3}\right|+C\end{array}$

Q21. Integrate the rational function : $\frac{1}{(e^x-1)}[\text{Hint}:\text{Put}{e}^x=t]$

Answer. $\begin{array}{c}\frac{1}{x(x^4-1)}\\ \text{Multiplying numerator and denominator by}{x}^3,\text{we obtain}\\ \frac{1}{x(x^4-1)}=\frac{x^3}{x^4(x^4-1)}\\ \therefore \int \frac{1}{x(x^4-1)}dx=\int \frac{x^3}{x^4(x^4-1)}dx\end{array}$ $\begin{array}{c}\text{Let}{x}^4=t\Rightarrow 4x^{3}dx=dt\\ \therefore \int \frac{1}{x(x^4-1)}dx=\frac{1}{4}\int \frac{dt}{t(t-1)}\end{array}$ $\begin{array}{c}\text{Let}\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{(t-1)}\\ 1=A(t-1)+Bt....\left(1\right)\\ \text{Substituting}t=0\text{and}1\text{in}\left(1\right),\text{we obtain}\\ A=-1\text{and}B=1\end{array}$ $\begin{array}{c}\Rightarrow \frac{1}{t(t+1)}=\frac{-1}{t}+\frac{1}{t-1}\\ \Rightarrow \int \frac{1}{x(x^4-1)}dx=\frac{1}{4}\int \{\frac{-1}{t}+\frac{1}{t-1}\}dt\end{array}$ $\begin{array}{cc}& =\frac{1}{4}[-\log |t|+\log |t-1|]+C\\ & =\frac{1}{4}\log \left|\frac{t-1}{t}\right|+C\\ & =\frac{1}{4}\log \left|\frac{x^4-1}{x^4}\right|+C\end{array}$

Q22. Choose the correct answer : $\int \frac{xdx}{(x-1)(x-2)}$ equals $\begin{array}{cc}\text{(A)}\log \left|\frac{(x-1{)}^2}{x-2}\right|+C& \text{(B)}\log \left|\frac{(x-2{)}^2}{x-1}\right|+C\\ \left(C\right)\log \left|{\left(\frac{x-1}{x-2}\right)}^2\right|+C& \text{(D)}\log |(x-1)(x-2)|+C\end{array}$

Answer. $\begin{array}{c}\frac{1}{(e^x-1)}\\ \text{Let}{e}^x=t\Rightarrow e^{x}dx=dt\\ \Rightarrow \int \frac{1}{e^x-1}dx=\int \frac{1}{t-1}\times \frac{dt}{t}=\int \frac{1}{t(t-1)}dt\end{array}$ $\begin{array}{c}\text{Let}\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}\\ 1=A(t-1)+Bt\\ \text{Substituting}t=1\text{and}t=0\text{in equation}\left(1\right),\text{we obtain}\end{array}$ $\begin{array}{cc}\therefore \frac{x}{(x-1)(x-2)}& =-\frac{1}{(x-1)}+\frac{2}{(x-2)}\\ \Rightarrow \int \frac{x}{(x-1)(x-2)}dx& =\int \{\frac{-1}{(x-1)}+\frac{2}{(x-2)}\}dx\\ & =-\log |x-1|+2\log |x-2|+C\\ & =\log \left|\frac{(x-2{)}^2}{x-1}\right|+C\end{array}$

Q23. Choose the correct answer : $\int \frac{dx}{x(x^2+1)}$ equals $\begin{array}{cc}\text{(A)}\log |x|-\frac{1}{2}\log (x^2+1)+C& \text{(B)}\log |x|+\frac{1}{2}\log (x^2+1)+C\\ \left(C\right)-\log |x|+\frac{1}{2}\log (x^2+1)+C& \text{(D)}\frac{1}{2}\log |x|+\log (x^2+1)+C\end{array}$

Answer. $\begin{array}{c}\text{Let}\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}\\ 1=A(x^2+1)+(Bx+C)x\\ \text{Equating the coefficients of}{x}^2,x,\text{and constant term, we obtain}\end{array}$ A + B = 0 C = 0 A = 1 On solving these equations, we obtain A = 1, B = −1, and C = 0 $\begin{array}{c}\therefore \frac{1}{x(x^2+1)}=\frac{1}{x}+\frac{-x}{x^2+1}\\ \Rightarrow \int \frac{1}{x(x^2+1)}dx=\int \{\frac{1}{x}-\frac{x}{x^2+1}\}dx\end{array}$ $\begin{array}{c}=\log |x|-\frac{1}{2}\log |x^2+1|+C\\ \text{Hence, the correct Answer is}A\end{array}$

Chapter-7 (integrals)

• Exercise 7.1
• Exercise 7.2
• Exercise 7.3
• Exercise 7.4
• Exercise 7.6
• Exercise 7.7
• Exercise 7.8
• Exercise 7.9
• Exercise 7.10
• Exercise 7.11