Mathematics

# $\int { \frac { x }{ { x }^{ 4 }-1 } } dx$

##### SOLUTION
$\displaystyle \int{\dfrac{x}{x^4-1}}dx$
$\Rightarrow u=x^2\Rightarrow du=2xdx$
$\displaystyle \int{\dfrac{1 du}{2(u^2-1)}}=\dfrac{1}{4}\displaystyle \int{\dfrac{(u+1)-(u-1)}{(u-1)(u+1)}}$
$=\dfrac{1}{4}\displaystyle \int{\dfrac{1}{u-1}-\dfrac{1}{u+1}}$
$=\dfrac{1}{4}\ln \left(\dfrac{u-1}{u+1}\right)=\dfrac{1}{4}\left(\dfrac{x^2-1}{x^2+1}\right)$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

Q1 Single Correct Medium
The set of values of 'a' which satisfy the equation $\displaystyle \int_{0}^{2}(t-log_2a)dt=log_2\left(\dfrac{4}{a^2}\right),$ is
• A. $a \in R$
• B. $a < 2$
• C. $a > 2$
• D. $a \in R^+$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate:
$\displaystyle \int \dfrac {dx}{x(x^{3}+8)}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
$\displaystyle \int_{0}^{\pi/2} \dfrac {\sin 2x}{1 + 2\cos^{2}x} dx$ is equal to
• A. $\dfrac {1}{2}\log 2$
• B. $\log 2$
• C. $\log 3$
• D. $\dfrac {1}{3}\log 3$
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1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
If $\displaystyle \int _{ 0 }^{ \pi /2 }{ \dfrac { \sin { x } }{ 1+\sin { x+\cos { x } } } }dx=\dfrac{\pi}{a}-\dfrac{1}{2}lnb,\ a,b \in N$ then
• A. $a+b=4$
• B. $a-b=4$
• C. $a-b=6$
• D. $a+b=6$

$\underset {n\rightarrow \infty}{lim}\dfrac{1^2+2^2+3^2+.....+n^2}{n^3}=.................$