#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 130 Maths Textbook Solution.

$\inline \frac{1}{3} \log |\cot x+1|-\frac{1}{6} \log \left|\cot ^{2} x-\cot x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+C$

Given: $\inline \int \frac{\cot x+\cot ^{3} x}{1+\cot ^{3} x} d x$

Hint: using partial fraction and $\inline \int \frac{1}{x} d x, \int \frac{1}{1+x^{2}} d x$

Explanation:

\inline \begin{aligned} &I=\int \frac{\cot x+\cot ^{3} x}{1+\cot ^{3} x} d x \\ &=\int \frac{\cot x\left(1+\cot ^{2} x\right)}{1+\cot ^{3} x} d x \\ &=\int \frac{\cot x \cos e c^{2} x}{1+\cot ^{3} x} d x \\ &=-\int \frac{t}{1+t^{3}} d t \end{aligned}                                                   $\inline \left[\begin{array}{l} p u t \cot x=t \\ -\cos e c^{2} x=d t \\ \cos e c^{2} x d x=-d t \end{array}\right]$

$\inline =-\int \frac{t}{(1+t)\left(t^{2}-t+1\right)}$

\inline \begin{aligned} &\text { now: } \frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{A}{1+t}+\frac{B}{t^{2}-t+1} \\ &\text { multiplying by }(1+t)\left(t^{2}-t+1\right) \\ &t=A\left(t^{2}-t+1\right)+(B t+C)(t+1) \end{aligned}

\begin{aligned} &\text { putting\: t }=-1 \\ &-1=A(1+1+1)+(B(-1)+C)(0) \\ &=3 A \Rightarrow A=-\frac{1}{3} \end{aligned}

\inline \begin{aligned} &\text { putting } t=0 \\ &0=A(0-0+1)+(B(0)+C)(0+1) \\ &0=A(1)+C(1) \\ &0=A+C \Rightarrow 0=\frac{-1}{3}+C \Rightarrow C=\frac{1}{3} \end{aligned}

\inline \begin{aligned} &\text { putting } t=+1 \\ &1=A(1-1+1)+(B+C)(2) \\ &1=A(1)+2 B+2 C \\ &1=\frac{-1}{3}+2 B+\frac{2}{3} \end{aligned}

\inline \begin{aligned} &1+\frac{1}{3}-\frac{2}{3}=2 B \\ &2 B=\frac{3+1-2}{3} \\ &2 B=\frac{2}{3} \Rightarrow B=\frac{1}{3} \end{aligned}

$\frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{\frac{-1}{3}}{t+1}+\frac{\frac{1}{3}(t+1)}{t^{2}-t+1}$

$\int \frac{t}{(1+t)\left(t^{2}-t+1\right)}=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{3} \int \frac{t+1}{t^{2}-t+1} d t$

$=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{3 \times 2} \int \frac{2 t+2-3+3}{t^{2}-t+1} d t$

$=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{6} \int \frac{2 t-1}{t^{2}-t+1} d t-\frac{1}{2} \int \frac{1}{t^{2}-t+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1}$

$=\frac{1}{3} \int \frac{1}{t+1} d t-\frac{1}{6} \int \frac{2 t-1}{t^{2}-t+1} d t-\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}}$

$=\frac{1}{3} \log |t+1|-\frac{1}{6} \log \left|t^{2}-t+1\right|-\frac{1}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C$

$=\frac{1}{3} \log |t+1|-\frac{1}{6} \log \left|t^{2}-t+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C$

$=\frac{1}{3} \log |\cot x+1|-\frac{1}{6} \log \left|\cot ^{2} x-\cot x+1\right|-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cot x-1}{\sqrt{3}}\right)+C$