Mathematics

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

##### SOLUTION

Consider the given integral.

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

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

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

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

$I=t-{{\tan }^{-1}}t+C$

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

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