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

The integral of (4+x^3)^2 is 16x+2x^4+(x^7)/7+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (4+x^3)^2

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

16 x + 2 x^{4} + \frac{x ^{7}}{7} + C

Solution steps

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

Expand

(4 + x^{3})^{2} : 16 + 8 x^{3} + x^{6}

= \int 16 + 8 x^{3} + x^{6} 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 16 d x + \int 8 x^{3} d x + \int x^{6} d x

\int 16 d x = 16 x

\int 8 x^{3} d x = 2 x^{4}

\int x^{6} d x = \frac{x ^{7}}{7}

= 16 x + 2 x^{4} + \frac{x ^{7}}{7}

Add a constant to the solution

= 16 x + 2 x^{4} + \frac{x ^{7}}{7} + C

\int (\frac{1}{x - 1})^{7} d x

\int sin (t) ln (cos (t)) d t

\int \frac{2 x + 3}{2 x + 1} d x

\int 1 - x^{4} x^{3} d x

What is the integral of (4+x^3)^2 ?
The integral of (4+x^3)^2 is 16x+2x^4+(x^7)/7+C