integral of x^3(5x^4+3)^2

解

\int x^{3} (5 x^{4} + 3)^{2} d x

\frac{25 x ^{12}}{12} + \frac{15 x ^{8}}{4} + \frac{9 x ^{4}}{4} + C

解答ステップ

\int x^{3} (5 x^{4} + 3)^{2} d x

因数

5 x^{4} + 3 : - (- 5 x^{4} - 3)

= \int x^{3} (- (- 5 x^{4} - 3))^{2} d x

(- (- 5 x^{4} - 3))^{2} = (- 5 x^{4} - 3)^{2}

= \int x^{3} (- 5 x^{4} - 3)^{2} d x

- 5 x^{4} - 3 = 3 (- \frac{5 x ^{4}}{3} - 1)

= \int x^{3} (3 (- \frac{5 x ^{4}}{3} - 1))^{2} d x

簡素化

x^{3} (3 (- \frac{5 x ^{4}}{3} - 1))^{2} : 25 x^{11} + 30 x^{7} + 9 x^{3}

= \int 25 x^{11} + 30 x^{7} + 9 x^{3} 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 25 x^{11} d x + \int 30 x^{7} d x + \int 9 x^{3} d x

\int 25 x^{11} d x = \frac{25 x ^{12}}{12}

\int 30 x^{7} d x = \frac{15 x ^{8}}{4}

\int 9 x^{3} d x = \frac{9 x ^{4}}{4}

= \frac{25 x ^{12}}{12} + \frac{15 x ^{8}}{4} + \frac{9 x ^{4}}{4}

定数を解答に追加する

= \frac{25 x ^{12}}{12} + \frac{15 x ^{8}}{4} + \frac{9 x ^{4}}{4} + C