Mathematics

# Integrals of sum particular function prove that $\displaystyle\int {{{dx} \over {{x^2} - {a^2}}} = {1 \over {2a}}\log \left| {{{x - a} \over {x + a}}} \right| + c}$

##### SOLUTION
Since $\dfrac{1}{x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=\dfrac{1}{2a}\left[\dfrac{1}{x-a}-\dfrac{1}{x+a}\right]$

Now, taking

$\displaystyle\int \dfrac{dx}{x^2-a^2}=\displaystyle\int \dfrac{1}{2a}\left[\dfrac{1}{x-a}-\dfrac{1}{x+a}\right]dx$

$=\dfrac{1}{2a}\left[\displaystyle\int \dfrac{dx}{x-a}-\displaystyle\int \dfrac{dx}{x+a}\right]$ [using formula $\displaystyle\int \dfrac{dx}{x}=log x$]

$=\dfrac{1}{2a}[log |x-a|-log|x+a|]+c$

$=\dfrac{1}{2a}log\left|\dfrac{x-a}{x+a}\right|+c$ proved.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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Q3 Assertion & Reason Hard
##### ASSERTION

The value of $\displaystyle \int_{0}^{\pi }xf\left ( \sin x \right )dx$ is $\displaystyle \frac{\pi }{2}\displaystyle \int_{0}^{\pi }f\left ( \sin x \right )dx$ or $\pi \displaystyle \int_{0}^{\pi /2}f\left ( \sin x \right )dx$

##### REASON

$\displaystyle \int_{0}^{a}f\left ( x \right )dx=\displaystyle \int_{0}^{a}f\left ( a-x \right )dx$ and $\displaystyle \int_{0}^{2a}f\left ( x \right )dx=2\displaystyle \int_{0}^{a}f\left ( x \right )dx$ If $f\left ( 2a-x \right )=f\left ( x \right )$

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