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  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.14 Question 4

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.14 Question 4

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Answer: x-\frac{5}{2} \tan ^{-1}\left|\frac{x}{2}\right|+c

Hint: To solve this integral, use special integral formula.

Given:  \int \frac{x^{2}-1}{x^{2}+4} d x

Solution:

Let,

\begin{aligned} &I=\int \frac{x^{2}-1}{x^{2}+4} d x=\int \frac{x^{2}-1+4-4}{x^{2}+4} d x\\ \\ &=\int \frac{\left(x^{2}+4\right)-1-4}{x^{2}+4} d x=\int\left(\frac{x^{2}+4}{x^{2}+4}-\frac{5}{x^{2}+4}\right) d x \\\\ &=\int\left(1-\frac{5}{x^{2}+4}\right) d x=\int 1 d x-5 \int \frac{1}{x^{2}+2^{2}} \end{aligned}

 

=x-5 \cdot \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+c \quad\quad\quad\quad\quad\quad\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left|\frac{x}{a}\right|+c\right]

 

=x-\frac{5}{2} \tan ^{-1}\left|\frac{x}{2}\right|+c

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