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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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