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explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 66

Answers (1)

Answer:

                \frac{1}{7} \log \frac{x-2}{x+2}+\frac{\sqrt{3}}{7} \tan ^{-1} \frac{x}{\sqrt{3}}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{x^{2}}{x^{4}-x^{2}-12} d x

Explanation:

Let

I=\int \frac{x^{2}}{x^{4}-x^{2}-12} d x

Let x^{2}= y

\begin{aligned} &\frac{y}{y^{2}-y-12}=\frac{y}{(y-4)(y+3)} \\ &\frac{y}{(y-4)(y+3)}=\frac{A}{y-4}+\frac{B}{y+3} \\ &y=A(y+3)+B(y-4) \end{aligned}

\begin{aligned} &y=-3 \\ &-3=-7 B \\ &B=\frac{3}{7} \\ &y=4 \\ &4=7 A \\ &A=\frac{4}{7} \end{aligned}

\begin{aligned} &\int \frac{x^{2}}{x^{4}-x^{2}-12}=\frac{4}{7} \int \frac{1}{x^{2}-4} d x+\frac{3}{7} \int \frac{1}{x^{2}+3} d x \\ &=\frac{1}{7} \log \frac{x-2}{x+2}+\frac{3}{7} \cdot \frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}+C \end{aligned}

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