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

The integral of (2-3x^2)^2 is sqrt(2/3)(2sqrt(6)x-2sqrt(6)x^3+(9sqrt(3)x^5)/(5sqrt(2)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2}{3} (26 x - 26 x^{3} + \frac{9 3 x ^{5}}{5 2}) + C

Solution steps

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

Apply Integral Substitution

= \int \frac{2}{3} (2 - 2 u^{2})^{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{3} \cdot \int (2 - 2 u^{2})^{2} d u

Expand

(2 - 2 u^{2})^{2} : 4 - 8 u^{2} + 4 u^{4}

= \frac{2}{3} \cdot \int 4 - 8 u^{2} + 4 u^{4} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{3} (\int 4 d u - \int 8 u^{2} d u + \int 4 u^{4} d u)

\int 4 d u = 4 u

\int 8 u^{2} d u = \frac{8 u ^{3}}{3}

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

= \frac{2}{3} (4 u - \frac{8 u ^{3}}{3} + \frac{4 u ^{5}}{5})

Substitute back

u = \frac{3}{2} x

= \frac{2}{3} 4 \cdot \frac{3}{2} x - \frac{8 ( \frac{3}{2} x ) ^{3}}{3} + \frac{4 ( \frac{3}{2} x ) ^{5}}{5}

Simplify

\frac{2}{3} 4 \cdot \frac{3}{2} x - \frac{8 ( \frac{3}{2} x ) ^{3}}{3} + \frac{4 ( \frac{3}{2} x ) ^{5}}{5} : \frac{2}{3} (26 x - 26 x^{3} + \frac{9 3 x ^{5}}{5 2})

= \frac{2}{3} (26 x - 26 x^{3} + \frac{9 3 x ^{5}}{5 2})

Add a constant to the solution

= \frac{2}{3} (26 x - 26 x^{3} + \frac{9 3 x ^{5}}{5 2}) + C

\int x^{2} - 3 x - 18 d x

\int tan^{2} (x) d x

\int \frac{ln ( q )}{q ^{2}} d q

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

\int [\frac{x ^{2} - 9}{x}] d x

What is the integral of (2-3x^2)^2 ?
The integral of (2-3x^2)^2 is sqrt(2/3)(2sqrt(6)x-2sqrt(6)x^3+(9sqrt(3)x^5)/(5sqrt(2)))+C