#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.20 Question 11 Maths Textbook Solution.

Answer:$x+\log \left|\frac{x}{x+1}\right|+c$

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

Hint: using Partial Fraction and $\int \frac{1}{x} d x$

Explanation: Let

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

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

$=1+\frac{x+1}{x(x-1)(x+1)}=1+\frac{1}{x(x-1)}$

$\therefore \int \frac{x^{3}+1}{x^{3}-x} d x=\int 1 d x+\int \frac{1}{x(x+1)} d x$

$=x+I_{1}$   ...................(1) Where $I_{1}=\int \frac{1}{x\left ( x+1 \right )}dx$

$\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{(x+1)}$

Multiplying by $x\left ( x+1 \right )$

$1=A\left ( x+1 \right )+B\left ( x+1 \right )$

Putting x = -1

$1=A\left ( 0+1 \right )+B\left ( 0 \right )\Rightarrow A=1$

\begin{aligned} &\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{(x+1)} \\ &\therefore \int \frac{1}{x(x+1)} d x=\int \frac{1}{x} d x-\int \frac{1}{x+1} d x \end{aligned}

$=\log |x|-\log |x+1|$

Put in (1)

\begin{aligned} I &=x+\log |x|-\log |x+1|+c \\ &=x+\log \left|\frac{x}{x+1}\right|+c \end{aligned}