Mathematics

# Integrate with respect to $x$:$\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx$

##### SOLUTION
We have,
$I=\displaystyle \int \dfrac {1}{x^{6}{(1+x^{-5})}^{\frac {1}{5}}}dx$

Let
$t=1+x^{-5}$

$\dfrac{dt}{dx}=0-\dfrac{5}{x^6}$

$\dfrac{dt}{5}=-\dfrac{dx}{x^6}$

Therefore,

$I=-\dfrac{1}{5}\displaystyle \int \dfrac{1}{t^{\frac{1}{5}}}dt$

$I=-\dfrac{1}{5}\dfrac{t^{-\frac{1}{5}+1}}{-\frac{1}{5}+1}+C$

$I=-\dfrac{1}{5}\dfrac{t^{\frac{4}{5}}}{\frac{4}{5}}+C$

$I=-\dfrac{t^{\frac{4}{5}}}{4}+C$

On putting the value of $t$, we get

$I=-\dfrac{(1+x^{-5})^{\frac{4}{5}}}{4}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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$\int e^x(\tan x+\sec^2 x)dx=e^x \tan x+C$

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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 $\displaystyle f(x)>0 \forall x \in R$