MCQ
જો $x({x^4} + 1)\phi (x) = 1,$ તો $\int_1^2 {\phi (x)\,dx = } $
- ✓$\frac{1}{4}\log \frac{{32}}{{17}}$
- B$\frac{1}{2}\log \frac{{32}}{{17}}$
- C$\frac{1}{4}\log \frac{{16}}{{17}}$
- Dએકપણ નહીં.
==> $\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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