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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 126 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 126 Maths Textbook Solution.

Answers (1)

Answer:2 \log |x|-\frac{1}{4} \log |x+1|-\frac{1}{4} \log |x-1|-\frac{3}{4} \log \left|x^{2}+1\right|+c

Given: \int \frac{x^{2}-2}{x^{5}-x} d x

Hint: using partial fraction

Explanation let I=\int \frac{x^{2}-2}{x^{5}-x} d x

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

Multiply by x(x+1)(x-1)\left(x^{2}+1\right)

x^{2}-2=A(x+1)(x-1)\left(x^{2}+1\right)+B(u)(u-1)\left(x^{2}+1\right)+C(x)(x+1)\left(x^{2}+1\right)+(D x+E)(x)(x+1)(x-1)

\begin{aligned} &\text { putting } x=0 \\ &0-2=A(1)(-1)(1)+B(0)+C(0)+(D x+6)(10) \\ &-2=-A \Rightarrow A=2 \end{aligned}

\begin{aligned} &\text { putting } x=1 \\ &1-2=A(2)(0)(2)+B(0)+C(1)(2)(2)+(D x+6)(0) \\ &-1=4 C=>C=\frac{-1}{4} \end{aligned}

\begin{aligned} &\text { putting } x=-1\\ &1-2=A(0)+B(-1)(-2)(2)+C(0)+D(0)\\ &-1=>4 B \Rightarrow B=\frac{-1}{4} \end{aligned}

\text { On solving }, D=-\frac{3}{2} \& E=0

\int \frac{x^{2}-2}{x^{5}-x} d x=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1} d x-\frac{3}{2} \int \frac{x}{x^{2}+1} d x

\int \frac{x^{2}-2}{x^{5}-x} d x=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1} d x-\frac{3}{2} \int \frac{x}{x^{2}+1} d x

=2 \int \frac{1}{x} d x-\frac{1}{4} \int \frac{1}{x+1} d x-\frac{1}{4} \int \frac{1}{x-1}-\frac{3}{4} \int \frac{2 x d x}{x^{2}+1}

=2 \log |x|-\frac{1}{4} \log |x+1|-\frac{1}{4} \log |x-1|-\frac{3}{4} \log \left|x^{2}+1\right|+c

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