What is the integral of x^3sqrt(x^2+10) ?

The integral of x^3sqrt(x^2+10) is 1/5 (x^2+10)^{5/2}-10/3 (x^2+10)^{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+10)

\int x^{3} x^{2} + 10 d x

\frac{1}{5} (x^{2} + 10)^{\frac{5}{2}} - \frac{10}{3} (x^{2} + 10)^{\frac{3}{2}} + C

Solution steps

\int x^{3} x^{2} + 10 d x

Apply u-substitution

: \int (u^{2} - 10) u^{2} d u

= \int (u^{2} - 10) u^{2} d u

Expand

(u^{2} - 10) u^{2} : u^{4} - 10 u^{2}

= \int u^{4} - 10 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 10 u^{2} d u

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

\int 10 u^{2} d u = \frac{10 u ^{3}}{3}

= \frac{u ^{5}}{5} - \frac{10 u ^{3}}{3}

Substitute back

u = x^{2} + 10

= \frac{( x ^{2} + 10 ) ^{5}}{5} - \frac{10 ( x ^{2} + 10 ) ^{3}}{3}

Simplify

\frac{( x ^{2} + 10 ) ^{5}}{5} - \frac{10 ( x ^{2} + 10 ) ^{3}}{3} : \frac{1}{5} (x^{2} + 10)^{\frac{5}{2}} - \frac{10}{3} (x^{2} + 10)^{\frac{3}{2}}

= \frac{1}{5} (x^{2} + 10)^{\frac{5}{2}} - \frac{10}{3} (x^{2} + 10)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{5} (x^{2} + 10)^{\frac{5}{2}} - \frac{10}{3} (x^{2} + 10)^{\frac{3}{2}} + C

\int \frac{x + 4}{x ( x + 2 ) ^{2}} d x

\int \frac{1}{cosh ( x ) - sinh ( x )} d x

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

\int \frac{x + 3}{( x ^{2} + 6 x ) ^{2}} d x

\int \frac{5 + 5 x}{1 + x ^{2}} d x

What is the integral of x^3sqrt(x^2+10) ?
The integral of x^3sqrt(x^2+10) is 1/5 (x^2+10)^{5/2}-10/3 (x^2+10)^{3/2}+C