integral of 6x^2arccos(x)

解

\int 6 x^{2} arccos (x) d x

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

解答ステップ

\int 6 x^{2} arccos (x) d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int x^{2} arccos (x) d x

部分積分を適用する

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

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

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

簡素化

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

定数を解答に追加する

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