Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# The value (s) of ∫01 [x4 (1 - x)4 / (1 + x2)] dx is (are)

1) [(22 / 7) – π]

2) 2 / 105

3) 0

4) [(71 / 15) – (3π / 2)]

Solution: (1) [(22 / 7) – π]

$$\begin{array}{l}\begin{array}{l} \text { Let } \mathrm{I}=\int_{0}^{1} \frac{\mathbf{x}^{4}(1-\mathrm{x})^{4}}{1+\mathrm{x}^{2}} \mathrm{dx} \\ =\int_{0}^{1} \frac{\left(\mathrm{x}^{4}-1\right)(1-\mathrm{x})^{4}+(1-\mathrm{x})^{4}}{\left(1+x^{2}\right)} \mathrm{dx} \\ =\int_{0}^{1}\left(\mathrm{x}^{2}-1\right)(1-\mathrm{x})^{4} \mathrm{~d} \mathrm{x}+\int_{0}^{1} \frac{\left(1+\mathrm{x}^{2}-2 \mathrm{x}\right)^{2}}{\left(1+\mathrm{x}^{2}\right)} \mathrm{dx} \\ =\int_{0}^{1}\left\{\left(\mathrm{x}^{2}-1\right)(1-\mathrm{x})^{4}+\left(1+\mathrm{x}^{2}\right)-4 \mathrm{x}+\frac{4 \mathrm{x}^{2}}{\left(1+\mathrm{x}^{2}\right)}\right\} \mathrm{d} \mathrm{x} \\ =\int_{0}^{1}\left(\left(\mathrm{x}^{2}-1\right)(1-\mathrm{x})^{4}+\left(1+\mathrm{x}^{2}\right)-4 \mathrm{x}+4 \frac{4}{1-\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x} \\ =\int_{0}^{1}\left(\mathrm{x}^{6}-4 \mathrm{x}^{5}+5 \mathrm{x}^{4}-4 \mathrm{x}^{2}+4-\frac{4}{1+\mathrm{x}^{2}}\right) \mathrm{d} \mathrm{x} \\ =\left[\frac{\mathrm{x}^{7}}{7}-\frac{4 \mathrm{x}^{6}}{6}+\frac{5 \mathrm{x}^{5}}{5}-\frac{4 \mathrm{x}^{3}}{3}+4 \mathrm{x}-4 \tan ^{-1} \mathrm{x}\right]_{0}^{1} \\ =\frac{1}{7}-\frac{4}{6}+\frac{5}{5}-\frac{4}{3}+4-4\left(\frac{\pi}{4}-0\right)\\=\frac{22}{7}-\pi \end{array}\end{array}$$