Mathematics

# The value of integratin $\displaystyle \int x\sec^{-1}xdx$ is

$\displaystyle \frac{1}{2}x^{2}\sec^{-1}x-\frac{1}{2}\sqrt{x^{2}-1}+c$

##### SOLUTION
The value of $\displaystyle \int x \sec^{-1}x dx$ is
$\displaystyle =\sec^{-1}x\cdot \dfrac {x^2}{2}-\displaystyle \int \dfrac {1}{\sqrt {x^2-1}}\cdot \dfrac {x^2}{2}dx$
$\displaystyle =\dfrac {1}{2}x^2\sec^{-1}x-\dfrac {1}{4}\displaystyle \int \dfrac {dx^2}{\sqrt {x^2-1}}dx$
$\displaystyle =\dfrac {1}{2}x^2\sec^{-1}x-\dfrac {1}{2}\sqrt {x^2-1}+c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 128

#### Realted Questions

Q1 Subjective Medium
Is there any formula like integral  $\int {{e^x}\left( {f\left( x \right)f'\left( x \right)} \right)dx} = {e^x}f\left( x \right) + C.$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Medium
The value of $\displaystyle \int_{-\pi/2}^{\pi/2} \dfrac{x^2cosx}{1+e^x}dx$ is equal to
• A. $\dfrac{\pi^2}{4}+2$
• B. $\pi^2-e^{{-\pi/2}}$
• C. $\pi^2+e^{{\pi/2}}$
• D. $\dfrac{\pi^2}{4}-2$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Solve :
$\displaystyle \int \cfrac{1}{x-x^{3}} dx$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve:
$\frac{1}{2}\int {\frac{{2x}}{{{{\left( {1 + x} \right)}^2}}}dx}$

Solve $\displaystyle\int \dfrac{x}{{{{\left( {x + 1} \right)}^2}}}dx$