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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 42

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 42

Answers (1)

Answer:

                \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3} x}{2}\right)-\tan ^{-1} x+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x \\

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x \\ &\frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{3 x^{2}+4} \\ &x^{2}=(A x+B)\left(3 x^{2}+4\right)+(C x+D)\left(x^{2}+1\right) \\ &x^{2}=(3 A+C) x^{3}+(3 B+D) x^{2}+(4 A+C) x+(4 B+D) \end{aligned}

Equating the similar term

\begin{aligned} &3 A+C=0 \\ &3 A=-C \end{aligned}                            (1)

4A+C= 0

4A-3A= 0                   [From equation (1)]

A= 0                               (2)

Equation (1)

\begin{aligned} &C=0 \\ &3 B+D=1 \end{aligned}                      (3)

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

Equation (3)

\begin{aligned} &3 B-4 B=1 \\ &-B=1 \\ &B=-1 \\ &D=4 \end{aligned}                               [From equation (4)]

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

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