What is the integral of x^3sqrt(x^2+32) ?

The integral of x^3sqrt(x^2+32) is 1/5 (x^2+32)^{5/2}-32/3 (x^2+32)^{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+32)

\int x^{3} x^{2} + 32 d x

\frac{1}{5} (x^{2} + 32)^{\frac{5}{2}} - \frac{32}{3} (x^{2} + 32)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 32 d x

Apply u-substitution

: \int (u^{2} - 32) u^{2} d u

= \int (u^{2} - 32) u^{2} d u

Expand

(u^{2} - 32) u^{2} : u^{4} - 32 u^{2}

= \int u^{4} - 32 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 32 u^{2} d u

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

\int 32 u^{2} d u = \frac{32 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{32 u ^{3}}{3}

Substitute back

u = x^{2} + 32

= \frac{( x ^{2} + 32 ) ^{5}}{5} - \frac{32 ( x ^{2} + 32 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 32 ) ^{5}}{5} - \frac{32 ( x ^{2} + 32 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 32)^{\frac{5}{2}} - \frac{32}{3} (x^{2} + 32)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 32)^{\frac{5}{2}} - \frac{32}{3} (x^{2} + 32)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 32)^{\frac{5}{2}} - \frac{32}{3} (x^{2} + 32)^{\frac{3}{2}} + C

\frac{d}{d x} (3 cos (x) + 2 sin (x))

\frac{d}{d x} (\frac{x + 1}{x ^{\frac{2}{3}} ( x + 3 ) ^{\frac{1}{3}}})

\int e^{k x} d x

\int t - \frac{1}{t} d t

(1, 5), (- 2, - 4)

What is the integral of x^3sqrt(x^2+32) ?
The integral of x^3sqrt(x^2+32) is 1/5 (x^2+32)^{5/2}-32/3 (x^2+32)^{3/2}+C