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Choose the correct answers in the value of $\int\limits_0^1\tan^{-1}\bigg(\frac{\large 2x-1}{\large 1+x-x^2}\bigg)dx\;is$\[(A)\;1\qquad(B)\;0\qquad(C)\;-1\qquad(D)\;\frac{\pi}{4}\]


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  • (i) $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x) dx$
Given $I=\int \limits_0^1 \tan ^{-1}\bigg(\frac{2x-1}{1+x-x^2}\bigg) dx$
2x-1 can be written as $x+x-1$ and
$1+x-x^2$ can be written as $1+x(1-x)$
Therefore $I=\int \limits_0^1 \tan ^{-1} \bigg(\frac{x-(1-x)}{1+x(1-x)}\bigg) dx$
This is of the form $\tan ^{-1}\bigg[\frac{x-y}{1+xy}\bigg]$
Therefore $I=\int \limits_0^1 \tan ^{-1}x-\tan ^{-1}(1-x) dx$ ------(1)
Now applying the property $ \int \limits_a^b f(x)dx=\int \limits_a^b f(a-x)dx$
$I=\int \limits_0^1 \tan ^{-1} (1-x)-\tan ^{-1} (1-1+x)dx$
$=\int \limits_0^1 \tan (1-x) -\tan^{-1}(x) dx $ ------(2)
Adding equ(1) and equ(2)
$2I=\int \limits_0^1 [\tan ^{-1}x -\tan ^{-1}(1-x)+\tan (1-x)-\tan ^{-1}x]dx$
Hence $I=0$
correct answer is B


answered Mar 14, 2013 by meena.p

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