integral of xsqrt(x^2+2x)

解

\int x x^{2} + 2 x d x

\frac{1}{2} x^{2} x^{2} + 2 x - \frac{1}{6} (- 3 x x^{2} + 2 x + (x^{2} + 2 x)^{\frac{3}{2}} - 3 ln x + x^{2} + 2 x + 1 + 3 x^{2} + 2 x) + C

解答ステップ

\int x x^{2} + 2 x d x

部分積分を適用する

= \frac{1}{2} x^{2} x^{2} + 2 x - \int \frac{x ^{2} ( x + 1 )}{2 x ^{2} + 2 x} d x

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

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

定数を解答に追加する

= \frac{1}{2} x^{2} x^{2} + 2 x - \frac{1}{6} (- 3 x x^{2} + 2 x + (x^{2} + 2 x)^{\frac{3}{2}} - 3 ln x + x^{2} + 2 x + 1 + 3 x^{2} + 2 x) + C