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
Enrolled Students 84

#### Realted Questions

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Evaluate $\int {\tan}^{-1}x dx$.

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Q2 Subjective Medium
Evaluate the definite integral   $\displaystyle \int_0^{\pi}\left (\sin^2\frac {x}{2}-\cos^2\frac {x}{2}\right )dx$

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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 $f\left ( 2a-x \right )=f\left ( x \right )$

• A. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• B. Assertion is correct but Reason is incorrect
• C. Both Assertion and Reason are incorrect
• D. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Hard
If $y(x)$ is the solution of the differential equation $\displaystyle (x + 2) \frac{dy}{dx} = x^2 + 4x - 9, x \neq -2$ and $y(0) = 0$, then $y(-4)$ is equal to:
• A. $1$
• B. $2$
• C. $-1$
• D. $0$

$\int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx}$