If x( x^4 + 1) (x) = 1, then _1^2 (x) ,dx = - Vidyadip

MCQ

If $x({x^4} + 1)\phi (x) = 1,$ then $\int_1^2 {\phi (x)\,dx = } $

✓
$\frac{1}{4}\log \frac{{32}}{{17}}$
B
$\frac{1}{2}\log \frac{{32}}{{17}}$
C
$\frac{1}{4}\log \frac{{16}}{{17}}$
D
None of these

✓

Answer

Correct option: A.

$\frac{1}{4}\log \frac{{32}}{{17}}$

a
(a) Here $\phi (x) = \frac{1}{{x({x^4} + 1)}} = \frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}$

==> $\int_1^2 {\phi (x)dx = \int_1^2 {\,\left( {\frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}} \right)} \,dx} $

$ = |\log x|_1^2 - \left| {\frac{1}{4}\log ({x^2} + 1)} \right|_1^2 = \frac{1}{4}\log \frac{{32}}{{17}}$.

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