NCERT Solutions Class 12 Maths (integrals) Miscellaneous Exercise

Q1. Integrate the function $\frac{1}{x-x^3}$

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

Q2. Integrate the function $\frac{1}{\sqrt{x+a}+\sqrt{(x+b)}}$

Answer. $\begin{array}{cc}\frac{1}{\sqrt{x+a}+\sqrt{x+b}}& =\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}\\ & =\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}\\ & =\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b}\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}}dx& =\frac{1}{a-b}\int (\sqrt{x+a}-\sqrt{x+b})dx\\ & =\frac{1}{(a-b)}[\frac{(x+a{)}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b{)}^{\frac{3}{2}}}{\frac{3}{2}}]\\ & =\frac{2}{3(a-b)}\left[\right(x+a{)}^{\frac{3}{2}}-(x+b{)}^{\frac{3}{2}}]+C\end{array}$

Q3. Integrate the function $\frac{1}{x\sqrt{ax-x^2}}\left[{\text{Hint: Put}}^{x=\frac{a}{t}}\right]$

Answer. $\begin{array}{c}\frac{1}{x\sqrt{ax-x^2}}\\ \text{Let}x=\frac{a}{t}\Rightarrow dx=-\frac{a}{t^2}dt\end{array}$ $\begin{array}{cc}\Rightarrow \int \frac{1}{x\sqrt{ax-x^2}}dx& =\int \frac{1}{\frac{a}{t}\sqrt{a\cdot \frac{a}{t}-{\left(\frac{a}{t}\right)}^2}}(-\frac{a}{t^2}dt)\\ & =-\int \frac{1}{at}\cdot \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{t^2}}}dt\end{array}$ $\begin{array}{cc}& =-\frac{1}{a}\int \frac{1}{\sqrt{\frac{t^2}{t}-\frac{t^2}{t^2}}}dt\\ & =-\frac{1}{a}\int \frac{1}{\sqrt{t-1}}dt\\ & =-\frac{1}{a}\left[2\sqrt{t-1}\right]+C\\ & =-\frac{1}{a}\left[2\sqrt{\frac{a}{x}-1}\right]+C\end{array}$ $\begin{array}{cc}& =-\frac{2}{a}\left(\frac{\sqrt{a-x}}{\sqrt{x}}\right)+C\\ & =-\frac{2}{a}\left(\sqrt{\frac{a-x}{x}}\right)+C\end{array}$

Q4. Integrate the function $\frac{1}{x^2{(x^4+1)}^{\frac{3}{4}}}$

Answer. $\begin{array}{c}\frac{1}{x^2{(x^4+1)}^{\frac{3}{4}}}\\ \text{Multiplying and dividing by}{x}^{-3}\text{, we obtain}\\ \frac{x^{-3}}{x^2\cdot x^{-3}{(x^4+1)}^{\frac{3}{4}}}=\frac{x^{-3}{(x^4+1)}^{\frac{-3}{4}}}{x^2\cdot x^{-3}}\end{array}$ $\begin{array}{cc}& =\frac{{(x^4+1)}^{\frac{-3}{4}}}{x^5\cdot {\left(x^4\right)}^{\frac{3}{4}}}\\ & =\frac{1}{x^5}{\left(\frac{x^4+1}{x^4}\right)}^{\frac{3}{4}}\\ =& \frac{1}{x^5}{(1+\frac{1}{x^4})}^{-\frac{3}{4}}\end{array}$ $\begin{array}{c}\text{Let}\frac{1}{x^4}=t\Rightarrow -\frac{4}{x^5}dx=dt\Rightarrow \frac{1}{x^5}dx=-\frac{dt}{4}\\ \therefore \int \frac{1}{x^2{(x^4+1)}^{\frac{3}{4}}}dx=\int \frac{1}{x^5}{(1+\frac{1}{x^4})}^{-\frac{3}{4}}dx\end{array}$ $\begin{array}{cc}& =-\frac{1}{4}\int (1+t{)}^{\frac{3}{4}}dt\\ & =-\frac{1}{4}\left[\frac{(1+t{)}^{\frac{1}{4}}}{\frac{1}{4}}\right]+C\end{array}$ $\begin{array}{c}=-\frac{1}{4}\frac{{(1+\frac{1}{x^4})}^{\frac{1}{4}}}{\frac{1}{4}}+C\\ =-{(1+\frac{1}{x^4})}^{\frac{1}{4}}+C\end{array}$

Q5. Integrate the function $\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}[\text{Hint:}\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})},\text{put}x=t^6]$

Answer. $\begin{array}{c}\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})}\\ \text{Let}x=t^6\Rightarrow dx=6t^{5}dt\end{array}$ $\begin{array}{cc}\therefore \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx& =\int \frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})}dx\\ & =\int \frac{6t^5}{t^2(1+t)}dt\\ & =6\int \frac{t^3}{(1+t)}dt\end{array}$ $\begin{array}{c}\text{On dividing, we obtain}\\ \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}dx=6\int \{(t^2-t+1)-\frac{1}{1+t}\}dt\end{array}$ $\begin{array}{c}=6[\left(\frac{t^3}{3}\right)-\left(\frac{t^2}{2}\right)+t-\log |1+t|]\\ =2x^{\frac{1}{2}}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\log (1+x^{\frac{1}{6}})+C\\ =2\sqrt{x}-3x^{\frac{1}{3}}+6x^{\frac{1}{6}}-6\log (1+x^{\frac{1}{6}})+C\end{array}$

Q6. Integrate the function $\frac{5x}{(x+1)(x^2+9)}$

Answer. $\begin{array}{c}\text{Let}\frac{5x}{(x+1)(x^2+9)}=\frac{A}{(x+1)}+\frac{Bx+C}{(x^2+9)}\dots .\left(1\right)\\ \Rightarrow 5x=A(x^2+9)+(Bx+C)(x+1)\\ \Rightarrow 5x=A{x}^2+9A+B{x}^2+Bx+Cx+C\end{array}$ $\begin{array}{c}\text{Equating the coefficients of}{x}^2,x,\text{and constant term, we obtain}\\ A+B=0\\ B+C=5\\ 9A+C=0\end{array}$ $\begin{array}{c}\text{On solving these equations, we obtain}\\ A=-\frac{1}{2},B=\frac{1}{2},\text{and}C=\frac{9}{2}\\ \text{From equation}\left(1\right),\text{we obtain}\end{array}$ $\frac{5x}{(x+1)(x^2+9)}=\frac{-1}{}$