MCQ
The value of $\int\limits_0^1 {\frac{{{x^4}{{(1 - x)}^4}}}{{1 + {x^2}}}dx} $ is equal to
- ✓$\frac{{22}}{7} - \pi $
- B$2$
- C$\frac {2}{105}$
- D$\frac{{71}}{{15}} - \frac{{3\pi }}{2}$
$=\int_{0}^{1} \frac{x^{4}\left(\left(1+x^{2}\right)-2 x\right)^{2}}{1+x^{2}} d x$
$=\int_{0}^{1} \frac{x^{4}\left(\left(1+x^{2}\right)^{2}+4 x^{2}-4 x \cdot\left(1+x^{2}\right)\right.}{1+x^{2}} d x$
$=\int_{0}^{1}\left(x^{4} \cdot\left(1+x^{2}\right)+\frac{4 x^{2}}{1+x^{2}}-4 x\right) d x=\frac{22}{7}-\pi$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.