What is the integral of (1+x^3)^2 ?

The integral of (1+x^3)^2 is x+(x^4)/2+(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 (1+x^3)^2

\int (1 + x^{3})^{2} d x

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

Solution steps

\int (1 + x^{3})^{2} d x

Expand

(1 + x^{3})^{2} : 1 + 2 x^{3} + x^{6}

= \int 1 + 2 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 1 d x + \int 2 x^{3} d x + \int x^{6} d x

\int 1 d x = x

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

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

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

Add a constant to the solution

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

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

\int 4 - 4 (36 x^{2}) d x

\int x^{- 3} e^{\frac{1}{x}} d x

\int cos (3 x + 10) d x

\int (x^{2} + 4 x) x^{3} + 6 x^{2} + 1 d x

What is the integral of (1+x^3)^2 ?
The integral of (1+x^3)^2 is x+(x^4)/2+(x^7)/7+C