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

The integral of x^2(3x^3-2)^4 is (27x^{15})/5-18x^{12}+24x^9-16x^6+(16x^3)/3+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{27 x ^{15}}{5} - 18 x^{12} + 24 x^{9} - 16 x^{6} + \frac{16 x ^{3}}{3} + C

Solution steps

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

Expand

x^{2} (3 x^{3} - 2)^{4} : 81 x^{14} - 216 x^{11} + 216 x^{8} - 96 x^{5} + 16 x^{2}

= \int 81 x^{14} - 216 x^{11} + 216 x^{8} - 96 x^{5} + 16 x^{2} 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 81 x^{14} d x - \int 216 x^{11} d x + \int 216 x^{8} d x - \int 96 x^{5} d x + \int 16 x^{2} d x

\int 81 x^{14} d x = \frac{27 x ^{15}}{5}

\int 216 x^{11} d x = 18 x^{12}

\int 216 x^{8} d x = 24 x^{9}

\int 96 x^{5} d x = 16 x^{6}

\int 16 x^{2} d x = \frac{16 x ^{3}}{3}

= \frac{27 x ^{15}}{5} - 18 x^{12} + 24 x^{9} - 16 x^{6} + \frac{16 x ^{3}}{3}

Add a constant to the solution

= \frac{27 x ^{15}}{5} - 18 x^{12} + 24 x^{9} - 16 x^{6} + \frac{16 x ^{3}}{3} + C

\frac{d}{d x} (k^{x} + x^{k})

\int_{0}^{2} 2 π (x + 1) (12 - 6 x) d x

g (x) = 3^{(\frac{x}{2})}

x \to 2 lim (\frac{( 3 )}{x - 2})

\int sec^{7} (3 x) tan (3 x) d x

What is the integral of x^2(3x^3-2)^4 ?
The integral of x^2(3x^3-2)^4 is (27x^{15})/5-18x^{12}+24x^9-16x^6+(16x^3)/3+C