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

MCQ

If $x\left(x^4+1\right) \phi(x)=1$, then $\int_1^2 \phi(x) d x=$

✓
$\frac{1}{4} \log \frac{32}{17}$
B
$\frac{1}{2} \log \frac{32}{17}$
C
$\frac{1}{4} \log \frac{16}{17}$
D
$\frac{1}{2} \log \frac{16}{17}$

✓

Answer

Correct option: A.

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

(A)
$\phi(x)=\frac{1}{x\left(x^4+1\right)}=\frac{1}{x}-\frac{x^3}{x^4+1}$
$\therefore \quad \int_1^2 \phi(x) d x=\int_1^2\left(\frac{1}{x}-\frac{x^3}{x^4+1}\right) d x$
$\begin{array}{l}=[\log x]_1^2-\left[\frac{1}{4} \log \left(x^4+1\right)\right]_1^2 \\ =\frac{1}{4} \log \frac{32}{17}\end{array}$

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