What is the integral of x^2(7-2x^3)^4 ?

The integral of x^2(7-2x^3)^4 is (2401x^3)/3-(1372x^6)/3+(392x^9)/3-(56x^{12})/3+(16x^{15})/(15)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^2(7-2x^3)^4

\int x^{2} (7 - 2 x^{3})^{4} d x

\frac{2401 x ^{3}}{3} - \frac{1372 x ^{6}}{3} + \frac{392 x ^{9}}{3} - \frac{56 x ^{12}}{3} + \frac{16 x ^{15}}{15} + C

Solution steps

\int x^{2} (7 - 2 x^{3})^{4} d x

Factor

7 - 2 x^{3} : - (- 7 + 2 x^{3})

= \int x^{2} (- (- 7 + 2 x^{3}))^{4} d x

(- (- 7 + 2 x^{3}))^{4} = (- 7 + 2 x^{3})^{4}

= \int x^{2} (- 7 + 2 x^{3})^{4} d x

Expand

x^{2} (- 7 + 2 x^{3})^{4} : 2401 x^{2} - 2744 x^{5} + 1176 x^{8} - 224 x^{11} + 16 x^{14}

= \int 2401 x^{2} - 2744 x^{5} + 1176 x^{8} - 224 x^{11} + 16 x^{14} d x

Apply the Sum Rule:

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

= \int 2401 x^{2} d x - \int 2744 x^{5} d x + \int 1176 x^{8} d x - \int 224 x^{11} d x + \int 16 x^{14} d x

\int 2401 x^{2} d x = \frac{2401 x ^{3}}{3}

\int 2744 x^{5} d x = \frac{1372 x ^{6}}{3}

\int 1176 x^{8} d x = \frac{392 x ^{9}}{3}

\int 224 x^{11} d x = \frac{56 x ^{12}}{3}

\int 16 x^{14} d x = \frac{16 x ^{15}}{15}

= \frac{2401 x ^{3}}{3} - \frac{1372 x ^{6}}{3} + \frac{392 x ^{9}}{3} - \frac{56 x ^{12}}{3} + \frac{16 x ^{15}}{15}

Add a constant to the solution

= \frac{2401 x ^{3}}{3} - \frac{1372 x ^{6}}{3} + \frac{392 x ^{9}}{3} - \frac{56 x ^{12}}{3} + \frac{16 x ^{15}}{15} + C

What is the integral of x^2(7-2x^3)^4 ?
The integral of x^2(7-2x^3)^4 is (2401x^3)/3-(1372x^6)/3+(392x^9)/3-(56x^{12})/3+(16x^{15})/(15)+C