integral of (x^2+1)^{3/2}

解

\int (x^{2} + 1)^{\frac{3}{2}} d x

\frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} ln x + x^{2} + 1 + \frac{3}{8} x x^{2} + 1 + C

解答ステップ

\int (x^{2} + 1)^{\frac{3}{2}} d x

部分積分を適用する

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

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

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

簡素化

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

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

定数を解答に追加する

= \frac{1}{4} x (x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} ln x + x^{2} + 1 + \frac{3}{8} x x^{2} + 1 + C