Mathematics

# Evaluate: $\displaystyle \int \dfrac{1}{(x + 1)\sqrt{x^2 - 1}}dx$

$\dfrac {\sqrt {x-1}}{\sqrt{x+1}}+c$

##### SOLUTION
$\displaystyle\int \dfrac{1}{(x-1)\sqrt{x^2-1}}dx$
Substitute $x=\sec u$
$dx=\sec u+\tan u du$
$=\dfrac{\sec u \tan u}{(\sec u+1)\sqrt{\sec^2y-1}}du$
$\sec^2u-1=\tan^2u$
$=\dfrac{\sec u}{\sec u+1}du=\displaystyle\int \left(\dfrac{\sec u+1}{\sec u+1}-\dfrac{1}{\sec u+1}\right)du$
$=\displaystyle\int \left(1-\dfrac{1}{\sec u +1}\right)du$
$=u-\displaystyle\int \dfrac{1}{\dfrac{\tan^2\dfrac{u}{2}+1}{1-\tan^2\dfrac{u}{2}}+1}du$
$v=\tan \dfrac{u}{2}$
$\Rightarrow du=\dfrac{2}{\sec^2\dfrac{u}{2}}$
$du=\dfrac{2}{v^2+1}dv$
$=u-\displaystyle\int \dfrac{(v-1)(v+1)}{v^2+1}dv$
$=u-\displaystyle\int \dfrac{v^2+1}{v^2+1}-\dfrac{2}{v^2+1}dv=u-\displaystyle\int 1-\dfrac{2}{v^2+1}dv$
$=u-(v-2\tan^{-1}(v))$
Resubstituting $v=\tan \dfrac{u}{2}$
$=u-\left(\tan \dfrac{u}{2}-2\tan^{-1}\left(\tan\dfrac{u}{2}\right)\right)$
$=\tan\dfrac{u}{2}$
Resubstituting $u=\sec^{1}(x)$
$\tan \left(\dfrac{\sec^{-1}(x)}{2}\right)=\dfrac{\sqrt{x-1}}{\sqrt{x+1}}$
$\therefore \displaystyle\int \dfrac{1}{(x+1)\sqrt{x^2-1}}dx=\dfrac{\sqrt{x-1}}{\sqrt{x+1}}+C$.

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

#### Realted Questions

Q1 Single Correct Medium
$\displaystyle\int e^x\cdot \dfrac{x}{(x+1)^2}dx$.
• A. $-e^{x}\dfrac{1}{x+1}$
• B. $e^{x}\dfrac{x}{x+1}$
• C. None of these
• D. $e^{x}\dfrac{1}{x+1}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
$\int _{ 0 }^{ a }{ \left( f(x)+f(-x) \right) } dx=$
• A. $-\int _{ -a }^{ a }{ f(x)dx }$
• B. $0$
• C. $-\int _{ -a }^{ a }{ f(-x)dxz }$
• D. $2\int _{ 0 }^{ a }{ f(x)dx }$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $I= \displaystyle \int_{1}^{2}\displaystyle \frac{dx}{\left ( x+1 \right )\sqrt{x^{2}-1}}$ then $I$ is equal to
• A. $1$
• B. $1/2$
• C. $1/\sqrt{2}$
• D. $1/\sqrt{3}$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Evaluate the definite integral   $\displaystyle \int_2^3\frac {dx}{x^2-1}$

Let $\displaystyle I_{1}=\int_{0}^{1}(1-x^{2})^{1/3} dx$  &  $\displaystyle I_{2}=\int_{0}^{1}(1-x^{3})^{1/2} dx$