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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 48

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 48

Answers (1)

best_answer

Answer:

                \frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x+C\right]

Hint:

            To solve this integration, we use partial fraction method   

Given:

           \int \frac{3}{(1-x)\left(1+x^{2}\right)} d x \\

Explanation:

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

Equating the similar terms

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

\begin{aligned} &B-C=0 \\ &B=C \end{aligned}                                  (2)

A+C= 3

B+B=3                           [From equation (1) and (2)]

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

Thus

\begin{aligned} &I=\frac{3}{2} \int \frac{1}{1-x} d x+\frac{3}{2} \int \frac{x}{1+x^{2}} d x+\frac{3}{2} \int \frac{1}{1+x^{2}} d x \\ &I=\frac{-3}{2} \log |1-x|+\frac{3}{4} \log \left|1+x^{2}\right|+\frac{3}{2} \tan ^{-1} x+C \\ &I=\frac{3}{4}\left[\log \left|\frac{1+x^{2}}{(1-x)^{2}}\right|+2 \tan ^{-1} x+C\right] \end{aligned}

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