Mathematics

# $\displaystyle \int{\dfrac{(x^{2}-1)dx}{(x^{2}+1)\sqrt{x^{4}+1}}}$ is equal to :

$None\ of\ these$

##### SOLUTION

We have,

$\int{\dfrac{\left( {{x}^{2}}-1 \right)dx}{\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{4}}+1}}}$

Now,

Let

$t=\dfrac{x}{\sqrt{{{x}^{4}}+1}}$

$dt=\dfrac{1-{{x}^{4}}}{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{2}}}}dx$

So,

$\int{\dfrac{{{x}^{2}}-1}{\left( {{x}^{2}}+1 \right)\sqrt{{{x}^{4}}+1}}dx=\int{\dfrac{\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)\left( {{x}^{4}}+1 \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{2}}}}}dx}$

$=\int{\dfrac{1}{1+2{{t}^{2}}}.\dfrac{{{x}^{4}}-1}{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{2}}}}}dx$

$=-\int{\dfrac{dt}{1+2{{t}^{2}}}}$

On integration and we get,

$-\int{\dfrac{dt}{1+2{{t}^{2}}}}=-\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\sqrt{2}t$

$=\dfrac{1}{\sqrt{2}}{{\tan }^{-1}}\dfrac{x\sqrt{2}}{\sqrt{{{x}^{4}}+1}}+C$

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

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