explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 40

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

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{2 x}{x^{3}-1} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{2 x}{x^{3}-1} d x \\ &I=\int \frac{2 x}{(x-1)\left(x^{2}+x+1\right)} d x \quad \quad \quad \quad \quad \quad \quad\left[a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right] \end{aligned}

\begin{aligned} &\frac{2 x}{(x+1)\left(x^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+x+1} \\ &2 x=A\left(x^{2}+x+1\right)+(B x+C)(x-1) \\ &2 x=x^{2}(A+B)+x(A-B+C)+(A-C) \end{aligned}

Equating the similar terms

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

\begin{aligned} &A-C=0 \\ &A=C \end{aligned}                          (2)

\begin{aligned} & A-B+C=2 \\ &A+A+A=2 \end{aligned}                 [From equation (1) and (2)]

\begin{aligned} &3 A=2 \\ &A=\frac{2}{3} \end{aligned}

Equation (1)

$B=\frac{-2}{3}$

Equation (2)

$C=\frac{2}{3}$
\begin{aligned} &\frac{2 x}{(x-1)\left(x^{2}+x+1\right)}=\frac{2}{3(x-1)}+\frac{\frac{-2}{3} x+\frac{2}{3}}{x^{2}+x+1} \\ &=\frac{2}{3(x-1)}+\frac{2-2 x}{3\left(x^{2}+2 x+1\right)} \end{aligned}
\begin{aligned} &I=\frac{2}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{2 x-2}{\left(x^{2}+x+1\right)} d x \\ &I=\frac{2}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{2 x+1}{x^{2}+x+1} d x-\frac{1}{3} \int \frac{-3 d x}{x^{2}+x+1} d x \\ &I=\frac{2}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{2 x+1}{x^{2}+x+1} d x+\int \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d x \end{aligned}
$I=\frac{2}{3} \log |x-1|-\frac{1}{3} \log \left|x^{2}+x+1\right|+\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$

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