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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 39

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 39

Answers (1)

Answer:

                \frac{1}{2} \log |x+1|-\frac{1}{2(x+1)}-\frac{1}{4} \log \left|x^{2}+1\right|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{1}{(x+1)^{2}\left(x^{2}+1\right)} d x \\

Explanation:

Let

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

\begin{aligned} &1=A(x+1)\left(x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)(x+1)^{2} \\ &1=(A+C) x^{3}+(A+B+2 C+D) x^{2}+(A+C+2 D) x+(A+B+D) \end{aligned}

Equating the similar terms

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

\begin{aligned} &A+B+2 C+D=0 \\ &A+B-2 A+D=0 \end{aligned}                      [From equation (1)]

-A+B+D=0               (2)

A+C+2D=0

2D=0                               [From equation (1)]

D=0                                  (3)

Equation (2)

\begin{aligned} &-A+B=0 \\ &A=B \end{aligned}                                  (4)

\begin{aligned} &A+B+D=1 \\ &A+A+0=1 \end{aligned}                    [From equation (4) and (3)]

\begin{aligned} &2 A=1 \\ &A=\frac{1}{2} \end{aligned}

Equation (4)

B=\frac{1}{2}

And equation (1)

C=\frac{-1}{2}

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

Posted by

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