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

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

Popular >

integral of 4x^3sqrt(x^2+2)

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

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

Solution steps

\int 4 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

= 4 \cdot \int x^{3} x^{2} + 2 d x

Apply u-substitution

: \int (u^{2} - 2) u^{2} d u

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

Expand

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

= 4 \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

= 4 (\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}

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

Substitute back

u = x^{2} + 2

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

Simplify

4 (\frac{( x ^{2} + 2 ) ^{5}}{5} - \frac{2 ( x ^{2} + 2 ) ^{3}}{3}) : 4 (\frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}})

= 4 (\frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}})

Add a constant to the solution

= 4 (\frac{1}{5} (x^{2} + 2)^{\frac{5}{2}} - \frac{2}{3} (x^{2} + 2)^{\frac{3}{2}}) + C

\int 9 ln (2 x) d x

\int x sin (x^{2}) cos (3 x^{2}) d x

\int cos (π) d x

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

\int \frac{5 e ^{x}}{1 - e ^{2 x}} d x

What is the integral of 4x^3sqrt(x^2+2) ?
The integral of 4x^3sqrt(x^2+2) is 4(1/5 (x^2+2)^{5/2}-2/3 (x^2+2)^{3/2})+C