MCQ
$\int_1^2 {\frac{{dx}}{{x(1 + {x^4})}}} =$
- A$\frac{1}{4}\log \frac{{17}}{{32}}$
- B$\frac{1}{4}\log \frac{{17}}{2}$
- C$\log \frac{{17}}{2}$
- ✓$\frac{1}{4}\log \frac{{32}}{{17}}$
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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