integral of (x^3)/(sqrt(4x^2+16))

解

\int \frac{x ^{3}}{4 x ^{2} + 16} d x

\frac{x ^{4}}{2 x ^{2} + 4} - \frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} + \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{16}{x ^{2} + 4} + C

解答ステップ

\int \frac{x ^{3}}{4 x ^{2} + 16} d x

部分積分を適用する

= \frac{x ^{4}}{2 x ^{2} + 4} - \int \frac{x ^{3} ( x ^{2} + 6 )}{( x ^{2} + 4 ) x ^{2} + 4} d x

\int \frac{x ^{3} ( x ^{2} + 6 )}{( x ^{2} + 4 ) x ^{2} + 4} d x = \frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} - 2 (\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{8}{x ^{2} + 4})

= \frac{x ^{4}}{2 x ^{2} + 4} - (\frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} - 2 (\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{8}{x ^{2} + 4}))

簡素化

\frac{x ^{4}}{2 x ^{2} + 4} - (\frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} - 2 (\frac{1}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{8}{x ^{2} + 4})) : \frac{x ^{4}}{2 x ^{2} + 4} - \frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} + \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{16}{x ^{2} + 4}

= \frac{x ^{4}}{2 x ^{2} + 4} - \frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} + \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{16}{x ^{2} + 4}

定数を解答に追加する

= \frac{x ^{4}}{2 x ^{2} + 4} - \frac{x ^{2} ( x ^{2} + 6 )}{x ^{2} + 4} + \frac{2}{3} (x^{2} + 4)^{\frac{3}{2}} - \frac{16}{x ^{2} + 4} + C