#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 123 Maths Textbook Solution.

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

Given: $\int \frac{x}{x^{3}-1} d x$

Hint: using partial function

Explanation let $I=\int \frac{x}{x^{3}-1} d x$

$\frac{x}{x^{3}-1}=\frac{x}{(x-1)\left(x^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B x+c}{x^{2}+x+1}$

Multiplying by  $(x-1)\left(x^{2}+x+1\right)$

$x=A\left(x^{2}+x+1\right)+(B x+C)(x-1)$

Putting x=1

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

Putting x=-1

\begin{aligned} &-1=A+(-2)(-B+C)\\ &-1=\frac{1}{3}+2 B-2 C\\ &2(B-C)=-1 \frac{-1}{3} \Rightarrow B-C=\frac{-2}{3}\\ &B-C=>\frac{-2}{3} \end{aligned}

Putting x=0

\begin{aligned} &0=A(0+0+1)+(0+C)(0-1) \\ &0=A(1)+C(-1) \\ &0=\frac{1}{3}-C=>C=\frac{1}{3} \end{aligned}

Put in (1)

\begin{aligned} &B-\frac{1}{3}=\frac{-2}{3}=>B=\frac{1}{3} \frac{-2}{3}=>B=\frac{-1}{3} \\ &\frac{x}{(x-1)\left(x^{2}+x+1\right)}=\frac{1}{3(x-1)}+\frac{\frac{-1}{3} x+\frac{1}{3}}{x^{2}+x+1} \\ &\int \frac{x d x}{(x-1)\left(x^{2}+x+1\right)}=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x-1}{x^{2}+x+1} d x \\ &=\frac{1}{3} \log |x-1|-\frac{1}{3} I_{1} \quad \text { (2) } \end{aligned}

Where $I_{1}=\int \frac{x-1}{x^{2}+x+1} d x$

\begin{aligned} &=\frac{1}{2} \int \frac{2(x-1)+3-3}{x^{2}+x+1} d x \\ &=\frac{1}{2} \int \frac{2 x+1}{x^{2}+x+1} d x-\frac{3}{2} \int \frac{1}{x^{2}+x+1} d x \\ &=\frac{1}{2} \int \frac{d t}{t}-\frac{3}{2} \int \frac{1}{x^{2}+x+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d x \quad\left[\begin{array}{l} x^{2}+x+1=t \\ (2 x+1) d x=d t \end{array}\right] \\ &=\frac{1}{2} \int \frac{d t}{t}-\frac{3}{2} \int \frac{1}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ &=\frac{1}{2} \log |t|-\frac{3}{2} \cdot \frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \\ &=\frac{1}{2} \log \left|x^{2}+x+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned}

By (2), we get

\begin{aligned} &=\frac{1}{3} \log |x-1|-\frac{1}{3}\left[\frac{1}{2} \log \left|x^{2}+x+1\right|-\sqrt{3} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]+C \\ &=\frac{1}{3} \log |x-1|-\frac{1}{6} \log \left|x^{2}+x+1\right|+\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+C \end{aligned}