The value of _1^2 dx x(1 + x^4 ) is - Vidyadip

MCQ

The value of $\int_1^2 {\frac{{dx}}{{x(1 + {x^4})}}} $ is

A
$\frac{1}{4}\log \frac{{17}}{{32}}$
B
$\frac{1}{4}\log \frac{{17}}{2}$
C
$\log \frac{{17}}{2}$
✓
$\frac{1}{4}\log \frac{{32}}{{17}}$

✓

Answer

Correct option: D.

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

d
(d) $\int_1^2 {\frac{{dx}}{{x(1 + {x^4})}} = \int_1^2 {\frac{{dx}}{{{x^5}\left( {1 + \frac{1}{{{x^4}}}} \right)}}} } $

Put $\left( {1 + \frac{1}{{{x^4}}}} \right) = z \Rightarrow \frac{{ - 4}}{{{x^5}}}dx = dz$

==> $\frac{{ - 1}}{4}\int_2^{17/16} {\frac{{dz}}{z} = \left[ {\frac{{ - 1}}{4}\log z} \right]_2^{17/16}} $

$= \frac{1}{4}\log 2 - \frac{1}{4}\log \frac{{17}}{{16}}$

==> $I = \frac{1}{4}\log \left( {\frac{{32}}{{17}}} \right)$.

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