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

The integral of x^3sqrt(a^2+x^2) is 1/15 (3x^2-2a^2)(x^2+a^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 x^3sqrt(a^2+x^2)

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

\frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}} + C

Solution steps

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

Simplify

x^{3} : x^{2} x

= \int x^{2} x a^{2} + x^{2} d x

Apply u-substitution

: \int \frac{u a ^{2} + u}{2} d u

= \int \frac{u a ^{2} + u}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int u a^{2} + u d u

Apply u-substitution

: \int (v - a^{2}) v d v

= \frac{1}{2} \cdot \int (v - a^{2}) v d v

Expand

(v - a^{2}) v : v^{\frac{3}{2}} - a^{2} v

= \frac{1}{2} \cdot \int v^{\frac{3}{2}} - a^{2} v d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int v^{\frac{3}{2}} d v - \int a^{2} v d v)

\int v^{\frac{3}{2}} d v = \frac{2}{5} v^{\frac{5}{2}}

\int a^{2} v d v = a^{2} \frac{2}{3} v^{\frac{3}{2}}

= \frac{1}{2} (\frac{2}{5} v^{\frac{5}{2}} - a^{2} \frac{2}{3} v^{\frac{3}{2}})

Substitute back

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

Simplify

\frac{1}{2} (\frac{2}{5} (a^{2} + x^{2})^{\frac{5}{2}} - a^{2} \frac{2}{3} (a^{2} + x^{2})^{\frac{3}{2}}) : \frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}}

= \frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{15} (3 x^{2} - 2 a^{2}) (x^{2} + a^{2})^{\frac{3}{2}} + C

\int y^{2} sin (2 y) d y

\int \frac{e ^{2 x}}{9 - 4 e ^{4 x}} d x

\int x (\frac{- 3}{x ^{3}} - x^{2} + \frac{x}{4}) d x

\int \frac{sin ( x ) - cot ( x )}{sin ^{2} ( x )} d x

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

What is the integral of x^3sqrt(a^2+x^2) ?
The integral of x^3sqrt(a^2+x^2) is 1/15 (3x^2-2a^2)(x^2+a^2)^{3/2}+C