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

解

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

\frac{3}{2} x (x^{2} + 4)^{\frac{1}{2}} + \frac{1}{4} x (x^{2} + 4)^{\frac{3}{2}} + 6 ln (\frac{1}{2} x + (x^{2} + 4)^{\frac{1}{2}}) + C

解答ステップ

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

簡素化

x^{2} : xx

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

部分積分を適用する

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

\int 3 x^{2} (x^{2} + 4)^{\frac{1}{2}} d x = 48 - \frac{1}{32} x (4 + x^{2})^{\frac{1}{2}} + 4 ln \frac{x + ( 4 + x ^{2} ) ^{\frac{1}{2}}}{2} + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}}

= x (x^{2} + 4)^{\frac{3}{2}} - 48 - \frac{1}{32} x (4 + x^{2})^{\frac{1}{2}} + 4 ln \frac{x + ( 4 + x ^{2} ) ^{\frac{1}{2}}}{2} + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}}

簡素化

x (x^{2} + 4)^{\frac{3}{2}} - 48 - \frac{1}{32} x (4 + x^{2})^{\frac{1}{2}} + 4 ln \frac{x + ( 4 + x ^{2} ) ^{\frac{1}{2}}}{2} + \frac{1}{64} x (x^{2} + 4)^{\frac{3}{2}} : \frac{3}{2} x (x^{2} + 4)^{\frac{1}{2}} + \frac{1}{4} x (x^{2} + 4)^{\frac{3}{2}} + 6 ln (\frac{1}{2} x + (x^{2} + 4)^{\frac{1}{2}})

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

定数を解答に追加する

= \frac{3}{2} x (x^{2} + 4)^{\frac{1}{2}} + \frac{1}{4} x (x^{2} + 4)^{\frac{3}{2}} + 6 ln (\frac{1}{2} x + (x^{2} + 4)^{\frac{1}{2}}) + C