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

The integral of (3-x^2)^2 is 9x-2x^3+(x^5)/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (3-x^2)^2

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

9 x - 2 x^{3} + \frac{x ^{5}}{5} + C

Solution steps

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

Expand

(3 - x^{2})^{2} : 9 - 6 x^{2} + x^{4}

= \int 9 - 6 x^{2} + x^{4} 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 9 d x - \int 6 x^{2} d x + \int x^{4} d x

\int 9 d x = 9 x

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

\int x^{4} d x = \frac{x ^{5}}{5}

= 9 x - 2 x^{3} + \frac{x ^{5}}{5}

Add a constant to the solution

= 9 x - 2 x^{3} + \frac{x ^{5}}{5} + C

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

\int \frac{x - 7}{x ( 2 x - 7 )} d x

\int \frac{2 x + 4}{( 3 x ^{2} + 12 x ) ^{5}} d x

\int cos (3 x) sin (5 x) d x

\int x^{4} + 3 x - 9 d x

What is the integral of (3-x^2)^2 ?
The integral of (3-x^2)^2 is 9x-2x^3+(x^5)/5+C