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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 47

$\frac{1}{8} \log x-\frac{1}{24} \log \left(x^{3}+8\right)+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{x\left(x^{3}+8\right)} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{x\left(x^{3}+8\right)} d x \\ &I=\int \frac{x^{2}}{x^{3}\left(x^{3}+8\right)} d x \end{aligned}                                       [Multiply and divide by$x^{2}$]

Let

\begin{aligned} &x^{3}=y \\ &3 x^{2} d x=d y \\ &x^{2} d x=\frac{d y}{3} \\ &I=\frac{1}{3} \int \frac{d y}{y(y+8)} \\ &\frac{1}{y(y+8)}=\frac{A}{y}+\frac{B}{y+8} \\ &1=A(y+8)+B y \\ &1=(A+B) y+8 A \end{aligned}

Equating both side

\begin{aligned} &8 A=1 \\ &A=\frac{1}{8} \\ &A+B=0 \\ &B=-A=\frac{-1}{8} \\ &\frac{1}{y(y+8)}=\frac{1}{8 y}-\frac{1}{8(y+8)} \end{aligned}

Thus

\begin{aligned} &I=\frac{1}{3} \int\left(\frac{1}{8 y}-\frac{1}{8(y+8)}\right) d y \\ &I=\frac{1}{24} \int \frac{1}{y} d y-\frac{1}{24} \int \frac{1}{y+8} d y \end{aligned}

\begin{aligned} &I=\frac{1}{24} \log |y|-\frac{1}{24} \log |y+8|+C \\ &I=\frac{1}{24} \log \left|x^{3}\right|-\frac{1}{24} \log \left|x^{3}+8\right|+C \\ &I=\frac{1}{8} \log |x|-\frac{1}{24} \log \left|x^{3}+8\right|+C \end{aligned}