[tex]\begin{align} \int x^7\sqrt{4-x^4} \text{ d}x &= \int (4-u)\sqrt u \left(-\dfrac{1}{4}\right) \text{ d}u \\ &= -\dfrac{1}{4}\int (4-u)u^{\frac{1}{2}} \text{ d}u \\ &= - \dfrac{1}{4} \int (4u^{\frac{1}{2}}-u^{\frac{3}{2}}) \text{ d}u \\ &= - \dfrac{1}{4}\left(\dfrac{4}{\frac{1}{2}+1}u^{\frac{1}{2}+1}-\dfrac{1}{\frac{3}{2}+1}u^{\frac{3}{2}+1}\right)+C \\ &= - \dfrac{1}{4}\left(\dfrac{8}{3}u^{\frac{3}{2}}-\dfrac{2}{5}u^{\frac{3}{2}}u^1\right)+C \\ &= - \dfrac{1}{4}u^{\frac{3}{2}}\left(\dfrac{8}{3}- \dfrac{2}{5}u\right)+C \\ &= - \dfrac{1}{4}(4-x^4)^{\frac{3}{2}}\left(\dfrac{8}{3}- \dfrac{2}{5}(4-x^4)\right)+C \\ &= - \dfrac{1}{4}(4-x^4)^{\frac{3}{2}}\left(\dfrac{8}{3}- \dfrac{8}{5}+\dfrac{2}{5}x^4\right)+C \\ &= - \dfrac{1}{4}(4-x^4)^{\frac{3}{2}}\left(\dfrac{16}{15}+\dfrac{6}{15}x^4\right)+C \\ &= - \dfrac{1}{4}(4-x^4)^{\frac{3}{2}}\left(\dfrac{2}{15}\right)(8+3x^4)+C \\ &= - \dfrac{1}{30}(3x^4+8)(4-x^4)^{\frac{3}{2}}+C \end{align} [/tex]
