integral of (x-1)^5+3(x-1)^2+5

解

\int (x - 1)^{5} + 3 (x - 1)^{2} + 5 d x

\frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x + C

解答ステップ

\int (x - 1)^{5} + 3 (x - 1)^{2} + 5 d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int (x - 1)^{5} d x + \int 3 (x - 1)^{2} d x + \int 5 d x

\int (x - 1)^{5} d x = \frac{( x - 1 ) ^{6}}{6}

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

\int 5 d x = 5 x

= \frac{( x - 1 ) ^{6}}{6} + x^{3} - 3 x^{2} + 3 x + 5 x

簡素化

\frac{( x - 1 ) ^{6}}{6} + x^{3} - 3 x^{2} + 3 x + 5 x : \frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x

= \frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x

定数を解答に追加する

= \frac{1}{6} (x - 1)^{6} + x^{3} - 3 x^{2} + 8 x + C