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

MCQ

The value of $\int_0^1 \frac{x^4+1}{x^2+1} d x$ 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)
$\int_0^1 \frac{x^4+1}{x^2+1} d x=\int_0^1 \frac{x^4-1}{x^2+1} d x+2 \int_0^1 \frac{d x}{x^2+1}$
$=\int_0^1\left(x^2-1\right) d x+2 \int_0^1 \frac{d x}{x^2+1}$
$=\left[\frac{x^3}{3}-x\right]_0^1+\left[2 \tan ^{-1} x\right]_0^1$
$\begin{array}{l}=-\frac{2}{3}+\frac{\pi}{2} \\ =\frac{3 \pi-4}{6}\end{array}$

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