Question

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Answer 1

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Let

$I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1} \: ({ \frac{1}{ {x}^{2} - x + 1} )} \, dx </p><p>$

$\implies \: I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1} \: ({ \frac{x + (1 - x)}{ 1 - x(1 - x)} )} \, dx </p><p>$

$\implies I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \: + \int\limits_{0}^{1} {tan}^{ - 1}{(1 - x)} \, dx$

We know that

$\displaystyle \int\limits_{0}^{a} f({x}) \, dx = \displaystyle \int\limits_{0}^{a} f(a - {x}) \, dx$

So

$\implies I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \: + \int\limits_{0}^{1} {tan}^{ - 1}{(1 - (1 - x))} \, dx$

$\implies I = 2\displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \:$

Again

$\displaystyle \int\ {tan}^{ - 1}{x} \, dx \:$

$= x {tan}^{ - 1} x - \frac{1}{2} log( {x}^{2} + 1)$

Hence

$\implies I = 2\displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \:$

$=2 \times \left. \ {x {tan}^{ - 1}x - \frac{1}{2} log( {x}^{2} + 1) } \right|_0^1$

$= 2( \frac{\pi}{4} - \frac{1}{2} log(2))$

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Answer 2

Let

$I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1} \: ({ \frac{1}{ {x}^{2} - x + 1} )} \, dxI=0∫1tan−1(x2−x+11)dx$

$\implies \: I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1} \: ({ \frac{x + (1 - x)}{ 1 - x(1 - x)} )} \, dx⟹I=0∫1tan−1(1−x(1−x)x+(1−x))dx$

$\implies I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \: + \int\limits_{0}^{1} {tan}^{ - 1}{(1 - x)} \, dx⟹I=0∫1tan−1xdx+0∫1tan−1(1−x)dx$

We know that

$\displaystyle \int\limits_{0}^{a} f({x}) \, dx = \displaystyle \int\limits_{0}^{a} f(a - {x}) \, dx0∫af(x)dx=0∫af(a−x)dx$

So

$\implies I = \displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \: + \int\limits_{0}^{1} {tan}^{ - 1}{(1 - (1 - x))} \, dx⟹I=0∫1tan−1xdx+0∫1tan−1(1−(1−x))dx$

$\implies I = 2\displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \:⟹I=20∫1tan−1xdx$

Again

$\displaystyle \int\ {tan}^{ - 1}{x} \, dx \:∫ tan−1xdx</p><p>= x {tan}^{ - 1} x - \frac{1}{2} log( {x}^{2} + 1)=xtan−1x−21log(x2+1)$

Hence

$\implies I = 2\displaystyle \int\limits_{0}^{1} {tan}^{ - 1}{x} \, dx \:⟹I=20∫1tan−1xdx$

=2 \times \left. \ {x {tan}^{ - 1}x - \frac{1}{2} log( {x}^{2} + 1) }

$\right|_0^1=2× xtan−1x−21log(x2+1)∣∣∣01$

$= 2( \frac{\pi}{4} - \frac{1}{2}$

$log(2))=2(4π−21log(2))$