Question

Evaluate:-∫ x^4+1/x^2+1 with limits 0 to 1

Answer 1

$\displaystyle\int\limits_{0}^{1}\left(\dfrac{x^4+1}{x^2+1}\right)dx=\int\limits_{0}^{1}\left(\dfrac{x^4-1+2}{x^2+1}\right)dx\\\\=\int\limits_{0}^{1}\left(\dfrac{x^4-1}{x^2+1}+\dfrac{2}{x^2+1}\right)dx\\\\=\int\limits_{0}^{1}\left(\dfrac{(x^2+1)(x^2-1)}{x^2+1}+\dfrac{2}{x^2+1}\right)dx\\\\=\int\limits_{0}^{1}\left(x^2-1+\dfrac{2}{x^2+1}\right)dx$

$\displaystyle=\int\limits_{0}^{1}x^2dx-\int\limits_{0}^{1}1dx+2\int\limits_{0}^{1}\left(\dfrac{1}{x^2+1}\right)dx\\\\=\dfrac{1}{3}\left[x^3\right]_{0}^{1}-\left[x\right]_{0}^{1}+2\left[\tan^{-1}x\right]_{0}^{1}$

$=\dfrac {1}{3}(1^3-0^3)-(1-0)+2(\tan^{-1}1-\tan^{-1}0)\\\\=\dfrac{1}{3}-1+\dfrac{\pi}{2}\\\\=\mathbf{\dfrac{3\pi-4}{6}}$