What is the integral of x^3sqrt(x^2+40) ?

The integral of x^3sqrt(x^2+40) is 1/5 (x^2+40)^{5/2}-40/3 (x^2+40)^{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 x^3sqrt(x^2+40)

\int x^{3} x^{2} + 40 d x

\frac{1}{5} (x^{2} + 40)^{\frac{5}{2}} - \frac{40}{3} (x^{2} + 40)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 40 d x

Apply u-substitution

: \int (u^{2} - 40) u^{2} d u

= \int (u^{2} - 40) u^{2} d u

Expand

(u^{2} - 40) u^{2} : u^{4} - 40 u^{2}

= \int u^{4} - 40 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

= \int u^{4} d u - \int 40 u^{2} d u

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

\int 40 u^{2} d u = \frac{40 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{40 u ^{3}}{3}

Substitute back

u = x^{2} + 40

= \frac{( x ^{2} + 40 ) ^{5}}{5} - \frac{40 ( x ^{2} + 40 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 40 ) ^{5}}{5} - \frac{40 ( x ^{2} + 40 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 40)^{\frac{5}{2}} - \frac{40}{3} (x^{2} + 40)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 40)^{\frac{5}{2}} - \frac{40}{3} (x^{2} + 40)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 40)^{\frac{5}{2}} - \frac{40}{3} (x^{2} + 40)^{\frac{3}{2}} + C

\frac{4}{x ^{2}}

\int \frac{( x - 95 )}{x ^{2} ( x + 5 )} d x

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

x \to 2 + lim (e^{\frac{3}{2 - x}})

\frac{d}{d x} (\frac{x ^{3} + y ^{3}}{x ^{2} + y ^{2}})

What is the integral of x^3sqrt(x^2+40) ?
The integral of x^3sqrt(x^2+40) is 1/5 (x^2+40)^{5/2}-40/3 (x^2+40)^{3/2}+C