Mathematics

# $\displaystyle \int\dfrac{t^{2}}{t^{3}+1} dt =$

##### SOLUTION

Consider the following Equation.

$I=\int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{{{t}^{3}}+1}}dt$

$x={{t}^{3}}+1$

$\dfrac{dx}{dt}=2{{t}^{2}}$

$=\int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}}{x}*\dfrac{1}{2{{t}^{2}}}}dx$

$=\dfrac{1}{2}\ln x+C$

$=\dfrac{1}{2}\ln \left( {{t}^{3}}+1 \right)+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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Q5 Passage Hard
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$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
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