Mathematics

Evaluate  :  $$\displaystyle \int \dfrac{\tan^{ - 1}x}{1+x^2}\ dx $$


SOLUTION
We have,
$$\displaystyle I=\int \dfrac{\tan^{ - 1}x}{1+x^2}\ dx $$

Let $$t=\tan^{-1}x$$

$$dt=\dfrac{1}{1+x^2}dx$$

Therefore,
$$\displaystyle I=\int t\ dt $$

$$I=\dfrac{t^2}{2}+C$$

$$I=\dfrac{(\tan^{-1}x)^2}{2}+C$$

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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