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

The integral of 2x^3sqrt(x^2-2) is 2(1/5 (x^2-2)^{5/2}+2/3 (x^2-2)^{3/2})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

2 (\frac{1}{5} (x^{2} - 2)^{\frac{5}{2}} + \frac{2}{3} (x^{2} - 2)^{\frac{3}{2}}) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x^{3} x^{2} - 2 d x

Apply u-substitution

= 2 \cdot \int (u^{2} + 2) u^{2} d u

Expand

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

= 2 \cdot \int u^{4} + 2 u^{2} 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

= 2 (\int u^{4} d u + \int 2 u^{2} d u)

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

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

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

Substitute back

u = x^{2} - 2

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

Simplify

2 (\frac{( x ^{2} - 2 ) ^{5}}{5} + \frac{2 ( x ^{2} - 2 ) ^{3}}{3}) : 2 (\frac{1}{5} (x^{2} - 2)^{\frac{5}{2}} + \frac{2}{3} (x^{2} - 2)^{\frac{3}{2}})

= 2 (\frac{1}{5} (x^{2} - 2)^{\frac{5}{2}} + \frac{2}{3} (x^{2} - 2)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{1}{5} (x^{2} - 2)^{\frac{5}{2}} + \frac{2}{3} (x^{2} - 2)^{\frac{3}{2}}) + C

What is the integral of 2x^3sqrt(x^2-2) ?
The integral of 2x^3sqrt(x^2-2) is 2(1/5 (x^2-2)^{5/2}+2/3 (x^2-2)^{3/2})+C