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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 44

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

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

Equating similar terms

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

$C=1$                              (2)

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

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

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