The value of _0^1 x^4 + 1 x^2 + 1 ,dx is - Vidyadip

MCQ

The value of $\int_0^1 {\frac{{{x^4} + 1}}{{{x^2} + 1}}\,dx} $ is

✓
$\frac{1}{6}(3\pi - 4)$
B
$\frac{1}{6}(3 - 4\pi )$
C
$\frac{1}{6}(3\pi + 4)$
D
$\frac{1}{6}(3 + 4\pi )$

✓

Answer

Correct option: A.

$\frac{1}{6}(3\pi - 4)$

a
(a) $I = \int_0^1 {\frac{{{x^4} + 1}}{{{x^2} + 1}}dx = \int_0^1 {\frac{{{x^4} - 1}}{{{x^2} + 1}}dx + 2\int_0^1 {\frac{{dx}}{{1 + {x^2}}}} } } $

==> $I = \int_0^1 {({x^2} - 1)} dx + 2\int_0^1 {\frac{{dx}}{{1 + {x^2}}}} $

==> $I = \left[ {\frac{{{x^3}}}{3} - x} \right]_0^1 + 2\,[{\tan ^{ - 1}}x]_0^1$

$ = - \frac{2}{3} + \frac{\pi }{2} = \frac{{(3\pi - 4)}}{6}$.

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