The value(s) of \int_{0}^{1} \frac{x^{4} \left(\left(\right. 1 - x \left.\right)\right)^{4}}{1 + x^{2}} \textrm{ } dx is(are)

Engineering

Mathematics

Properties of Definite Integral

Question

The value(s) of $\int_{0}^{1} \frac{x^{4} {(1 - x)}^{4}}{1 + x^{2}} dx$ is(are)

$\frac{71}{15} - \frac{3 π}{2}$

$\frac{2}{105}$

$\frac{22}{7} - π$

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$\int_{0}^{1} \frac{x^{4} {(1 - 2 x + x^{2})}^{2}}{1 + x^{2}} dx$

N^r = x⁴{(1 + x²)² + 4x² – 4x(1 + x²)}

$\int_{0}^{1} \frac{x^{4} {{(1 + x^{2})}^{2} + 4 x^{2} - 4 x (1 + x^{2})}}{1 + x^{2}} dx = \int_{0}^{1} (x^{4} (1 + x^{2}) + \frac{4 x^{6}}{1 + x^{2}} - 4 x^{5}) dx$

$= \int_{0}^{1} (x^{4} + x^{6} + \frac{4 (x^{6} + 1 - 1)}{1 + x^{2}} - 4 x^{5}) dx = \int_{0}^{1} (x^{4} + x^{6} + 4 (x^{4} - x^{2} + 1) - \frac{4}{1 + x^{2}} - 4 x^{5}) dx$

$= \int_{0}^{1} (5 x^{4} + x^{6} - 4 x^{5} - 4 x^{2} + 4 - \frac{4}{1 + x^{2}}) dx = {(x^{5} + \frac{x^{7}}{7} - \frac{4 x^{6}}{6} - \frac{4 x^{3}}{3} + 4 x - 4 \tan^{- 1} x)}_{0}^{1}$

$1 + \frac{1}{7} - \frac{2}{3} - \frac{4}{3} + 4 - 4 \cdot \frac{π}{4} = 5 + \frac{1}{7} - \frac{6}{3} - π = \frac{22}{7} - π$