explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 46

$\frac{1}{4} \log \left|\frac{x^{4}}{x^{4}+1}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{x\left(x^{4}+1\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{x\left(x^{4}+1\right)} d x \\ &I=\int \frac{x^{3}}{x^{4}\left(x^{4}+1\right)} d x \end{aligned}                  [Multiply and divide by $x^{3}$]

Let

\begin{aligned} &x^{4}=y \\ &4 x^{3} d x=d y \\ &x^{3} d x=\frac{d y}{4} \\ &I=\frac{1}{4} \int \frac{d y}{y(y+1)} \end{aligned}

Now

\begin{aligned} &\frac{1}{y(y+1)}=\frac{A}{y}+\frac{B}{y+1} \\ &1=A(y+1)+B y \\ &1=(A+B) y+A \end{aligned}

Equating similar terms

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

\begin{aligned} &\frac{1}{y(y+1)}=\frac{1}{y}+\frac{(-1)}{y+1} \\ &I=\frac{1}{4} \int\left(\frac{1}{y}\right) d y+\frac{1}{4} \int \frac{-1}{y+1} d y \\ &I=\frac{1}{4} \log |y|+\left(\frac{-1}{4}\right) \log |y+1|+C \end{aligned}

\begin{aligned} &I=\frac{1}{4} \log \left|\frac{y}{y+1}\right|+C \\ &I=\frac{1}{4} \log \left|\frac{x^{4}}{x^{4}+1}\right|+C \quad\quad\quad\quad\quad\quad\left[y=x^{4}\right] \end{aligned}