Mathematics

# $\displaystyle \int{ { e }^{ 3\log { x } }{ \left( { x }^{ 4 }+1 \right) }^{ -1 }dx }$ is equal to ?

$\dfrac { 1 }{ 4 } \log { \left( { x }^{ 4 }+1 \right) } +C$

##### SOLUTION
$\int e^{3 \log x} (x^{4}+1)^{-1} dx= \int e^{\log x^{3}} (x^{2}+1)^{-1} dx$
$= \int \dfrac{x^{3}}{1+x^{4}} dx$
Let $14x^{4} =t$
$x^{3} dx =dt$
$\Rightarrow = \int \dfrac{dt}{4t}$
$=\dfrac{1}{4} \log t+c$
$=\dfrac{1}{4} \log (1+x^{4})+c$

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

#### Realted Questions

Q1 Subjective Medium
$\int {\frac{1}{{\left( {x - 1} \right)\left( {x - 2} \right)}}dx}$

1 Verified Answer | Published on 17th 09, 2020

Q2 One Word Medium
$\displaystyle \int_{0}^{\infty }\frac{dx}{\left ( x+\sqrt{1+x^{2}} \right )^{n}} = f(n)$
Find the value if $n=6$, can be expressed as $a/b$ in simplest form, then $b-a = ?$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard
If $\displaystyle f (x) =\int_{0}^{\pi}\frac{t \sin t \:dt}{\sqrt{1+\tan ^2 x\sin ^2 t }}$ for $0 < x < \frac{\pi}{2}$, then
• A. $\displaystyle f(0^+)= -\pi$
• B. $\displaystyle f(\frac{\pi}{4})=\frac{\pi^2}{8}$
• C. f is continuous but not differentiable in $\left ( 0,\frac{\pi}{2} \right )$
• D. f is continuous and differentiable in $\left ( 0,\frac{\pi}{2} \right )$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
Antiderivation of $\dfrac {\sin^{2}x}{1+\sin^{2}x}$ w.r.t is:
• A. $x-\dfrac {2}{\sqrt {2}}arc\tan(\sqrt {2}\tan x)+C$
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• C. $x-\sqrt {2} arc\tan(\sqrt {2}\tan x)+C$
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Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$