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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 43

$\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x \\$

Explanation:

Let

\begin{aligned} &I=\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x \\ &I=\int \frac{3 x+5}{x^{2}(x-1)-1(x-1)} d x \\ &I=\int \frac{3 x+5}{\left(x^{2}-1\right)(x-1)} d x \end{aligned}

\begin{aligned} &I=\int \frac{3 x+5}{(x+1)(x-1)^{2}} d x \\ &\frac{3 x+5}{(x+1)(x-1)^{2}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1} \\ &3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2} \end{aligned}           (1)

Put $x= 1$in equation (1)

\begin{aligned} &3+5=A(0)+B(2)+C(0) \\ &8=2 B \\ &B=4 \end{aligned}

Put $x= -1$ in equation (1)

\begin{aligned} &-3+5=A(0)+B(0)+C(4) \\ &2=4 C \\ &C=\frac{1}{2} \end{aligned}

Put $x=0$in equation (1)

\begin{aligned} &5=A(-1)(1)+B+C \\ &5=-A+B+C \\ &5=-A+4+\frac{1}{2} \\ &1=-A+\frac{1}{2} \end{aligned}

\begin{aligned} &\frac{1}{2}=-A \\ &A=\frac{-1}{2} \\ &\frac{3 x+5}{(x+1)(x-1)^{2}}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^{2}}+\frac{1}{2(x+1)} \end{aligned}

Thus
\begin{aligned} &I=\frac{-1}{2} \int \frac{d x}{x-1}+4 \int \frac{d x}{(x-1)^{2}}+\frac{1}{2} \int \frac{d x}{x+1} \\ &I=\frac{-1}{2} \log |x-1|+(-4) \frac{1}{x-1}+\frac{1}{2} \log |x+1|+C \\ &I=\frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+C \end{aligned}