What is the integral of x^3sqrt(x^2+31) ?

The integral of x^3sqrt(x^2+31) is 1/5 (x^2+31)^{5/2}-31/3 (x^2+31)^{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+31)

\int x^{3} x^{2} + 31 d x

\frac{1}{5} (x^{2} + 31)^{\frac{5}{2}} - \frac{31}{3} (x^{2} + 31)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 31 d x

Apply u-substitution

: \int (u^{2} - 31) u^{2} d u

= \int (u^{2} - 31) u^{2} d u

Expand

(u^{2} - 31) u^{2} : u^{4} - 31 u^{2}

= \int u^{4} - 31 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 31 u^{2} d u

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

\int 31 u^{2} d u = \frac{31 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{31 u ^{3}}{3}

Substitute back

u = x^{2} + 31

= \frac{( x ^{2} + 31 ) ^{5}}{5} - \frac{31 ( x ^{2} + 31 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 31 ) ^{5}}{5} - \frac{31 ( x ^{2} + 31 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 31)^{\frac{5}{2}} - \frac{31}{3} (x^{2} + 31)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 31)^{\frac{5}{2}} - \frac{31}{3} (x^{2} + 31)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 31)^{\frac{5}{2}} - \frac{31}{3} (x^{2} + 31)^{\frac{3}{2}} + C

x \to 0 lim (I n (x))

\int_{- 2}^{3} (x - 6) - (x^{2} - 12) d x

x \to 1 lim (\frac{8}{x ^{3} - 1})

x \to 0 lim (7 x)

\frac{d}{d x} (ln (\frac{x + a}{x - a}))

What is the integral of x^3sqrt(x^2+31) ?
The integral of x^3sqrt(x^2+31) is 1/5 (x^2+31)^{5/2}-31/3 (x^2+31)^{3/2}+C