What is the integral of x^3(x^4+5)^3 ?

The integral of x^3(x^4+5)^3 is (x^{16}}{16}+\frac{5x^{12})/4+(75x^8)/8+(125x^4)/4+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{x ^{16}}{16} + \frac{5 x ^{12}}{4} + \frac{75 x ^{8}}{8} + \frac{125 x ^{4}}{4} + C

Solution steps

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

Expand

x^{3} (x^{4} + 5)^{3} : x^{15} + 15 x^{11} + 75 x^{7} + 125 x^{3}

= \int x^{15} + 15 x^{11} + 75 x^{7} + 125 x^{3} 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 x^{15} d x + \int 15 x^{11} d x + \int 75 x^{7} d x + \int 125 x^{3} d x

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

\int 15 x^{11} d x = \frac{5 x ^{12}}{4}

\int 75 x^{7} d x = \frac{75 x ^{8}}{8}

\int 125 x^{3} d x = \frac{125 x ^{4}}{4}

= \frac{x ^{16}}{16} + \frac{5 x ^{12}}{4} + \frac{75 x ^{8}}{8} + \frac{125 x ^{4}}{4}

Add a constant to the solution

= \frac{x ^{16}}{16} + \frac{5 x ^{12}}{4} + \frac{75 x ^{8}}{8} + \frac{125 x ^{4}}{4} + C

\int \frac{2 + 3 x}{x - 3} d x

\int x^{2} + 6 x + 9 d x

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

\int \frac{( 8 x ^{2} + 3 )}{x} d x

\int \frac{1}{7 s - 1} d s

What is the integral of x^3(x^4+5)^3 ?
The integral of x^3(x^4+5)^3 is (x^{16}}{16}+\frac{5x^{12})/4+(75x^8)/8+(125x^4)/4+C