Mathematics

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


SOLUTION

Consider the given integral.

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

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

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

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

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

 

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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