Mathematics

$$I=\int_{0}^{1}{\tan^{-1}(\frac{2x-1}{1+x-x^{2}})}$$


SOLUTION
$$I=\displaystyle \int _0^1 \tan^{-1} \left (\dfrac {2x-1}{1+x-x^2} \right)dx$$
$$Hx-x^2=t$$
$$(1-2n)dx=dt$$
$$=\dfrac {1}{1+\left (\dfrac {2n-1}{1-x-x^2}\right)^2}\displaystyle \int _0^1 +C$$
$$\Rightarrow \ \left [\dfrac {1}{1+\left (\dfrac {2(1)-1}{1-1-1}\right)^2} -\dfrac {1}{1+\left (\dfrac {2(0)-1}{1-0-0}\right)^2}  \right]+C$$
$$\Rightarrow \ \left [\dfrac {1}{1+\left (\dfrac {2-1}{-1}\right)^2} -\dfrac {1}{1-\left (\dfrac {0-1}{1}\right)^2} \right]+C$$
$$=\left [\dfrac {1}{1+1} -\dfrac {1}{1-1}\right]$$
$$\Rightarrow \ \dfrac {1}{2}+C$$

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Subjective Medium Published on 17th 09, 2020
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