integral of (4x^2+1)^{3/2}

解

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

\frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1 + C

解答ステップ

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

部分積分を適用する

= x (4 x^{2} + 1)^{\frac{3}{2}} - \int 12 x^{2} (4 x^{2} + 1)^{\frac{1}{2}} d x

\int 12 x^{2} (4 x^{2} + 1)^{\frac{1}{2}} d x = \frac{3}{16} (4 x (4 x^{2} + 1)^{\frac{3}{2}} - 2 x 4 x^{2} + 1 - ln 2 x + 4 x^{2} + 1)

= x (4 x^{2} + 1)^{\frac{3}{2}} - \frac{3}{16} (4 x (4 x^{2} + 1)^{\frac{3}{2}} - 2 x 4 x^{2} + 1 - ln 2 x + 4 x^{2} + 1)

簡素化

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

= \frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1

定数を解答に追加する

= \frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1 + C