The integral _0^1 d x 1-x+x^2 has the value - Vidyadip

MCQ

The integral $\int_0^1 \frac{d x}{1-x+x^2}$ has the value

A
$\frac{2 \pi}{3}$
B
$\frac{2 \pi}{\sqrt{3}}$
✓
$\frac{2 \pi}{3 \sqrt{3}}$
D
$\frac{4 \pi}{3}$

✓

Answer

Correct option: C.

$\frac{2 \pi}{3 \sqrt{3}}$

(C)
$\int_0^1 \frac{d x}{x^2-x+1}=\int_0^1 \frac{d x}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}$
$=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x-1}{\sqrt{3}}\right)\right]_0^1$
$=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right]$
$=\frac{2}{\sqrt{3}}\left[\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)\right]=\frac{2}{\sqrt{3}} \cdot \frac{\pi}{3}=\frac{2 \pi}{3 \sqrt{3}}$

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