Mathematics

$\int x^{3}d(tan^{-1}x)$ is equal to

SOLUTION

We have,

$I=\int{{{x}^{3}}{{\tan }^{-1}}xdx}$

Then,

On integration by parts and we get,

$I=\int{{{x}^{3}}{{\tan }^{-1}}xdx}$

$\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\downarrow$

$I=\,\,\,\,\,v.\,u$

Using, ILATE and we know that,

$I=\int{u.vdx}$

$I=u\int{vdx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}dx}$

Then,

$I={{\tan }^{-1}}x\int{{{x}^{3}}dx-\int{\left( \dfrac{d}{dx}{{\tan }^{-1}}x\int{{{x}^{3}}dx} \right)dx}}+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\int{\dfrac{1}{1+{{x}^{2}}}\dfrac{{{x}^{4}}}{4}dx}+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\int{\dfrac{{{x}^{4}}}{1+{{x}^{2}}}dx}+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\int{\dfrac{{{x}^{4}}-1+1}{1+{{x}^{2}}}dx}+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\int{\left( \dfrac{{{x}^{4}}-1}{1+{{x}^{2}}}+\dfrac{1}{1+{{x}^{2}}} \right)dx}+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\int{\left( {{x}^{2}}-1 \right)dx+\int{\dfrac{1}{1+{{x}^{2}}}}dx}+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\int{\left( {{x}^{2}}-1 \right)dx+{{\tan }^{-1}}x+C}$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\int{{{x}^{2}}dx+\dfrac{1}{4}\int{1}dx+{{\tan }^{-1}}x+C}$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{1}{4}\times \dfrac{{{x}^{3}}}{3}+\dfrac{1}{4}x+{{\tan }^{-1}}x+C$

$I=\dfrac{{{x}^{4}}}{4}{{\tan }^{-1}}x-\dfrac{{{x}^{3}}}{12}+\dfrac{1}{4}x+{{\tan }^{-1}}x+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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