∫01 tan−1⁡1x2−x+1 is equal to - Infinity Learn by Sri Chaitanya

A
log 2
B
-log 2
C
$\frac{π}{2} + \log 2$
D
$\frac{π}{2} - \log 2$

Solution:

$\begin{aligned} \int_{0}^{1} \tan^{- 1} (\frac{1}{x^{2} - x + 1}) dx \\ = \int_{0}^{1} \tan^{- 1} \{\frac{x - (x - 1)}{1 + x (x - 1)}\} \end{aligned}$

$\begin{array}{l} = \int_{0}^{1} \tan^{- 1} xdx - \int_{0}^{1} \tan^{- 1} (x - 1) dx \\ = \int_{0}^{1} \tan^{- 1} xdx + \int_{0}^{1} \tan^{- 1} (1 - 1 + x) dx \\ = \int_{0}^{1} \tan^{- 1} xdx + \int_{0}^{1} \tan^{- 1} xdx \\ = 2 \int_{0}^{1} \tan^{- 1} xdx \\ = 2 {[{xtan}^{- 1} x - \int \frac{x}{1 + x^{2}} dx]}_{0}^{1} \\ = 2 {[{xtan}^{- 1} x - \frac{1}{2} \log (1 + x^{2})]}_{0}^{1} \\ = 2 [\{1 \tan^{- 1} 1 - \frac{1}{2} \log (2)\} - \{0 - \frac{1}{2} \log 1\}] \\ = 2 [\frac{π}{4} - \frac{1}{2} \log 2 - 0] = \frac{π}{2} - \log 2 \end{array}$

$\int_{0}^{1} \tan^{- 1} (\frac{1}{x^{2} - x + 1})$ is equal to

Solution:

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