What is the integral of x^3sqrt(1+16x^2) ?

The integral of x^3sqrt(1+16x^2) is 1/1920 (24x^2-1)(16x^2+1)^{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(1+16x^2)

\int x^{3} 1 + 16 x^{2} d x

\frac{1}{1920} (24 x^{2} - 1) (16 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 1 + 16 x^{2} d x

Apply u-substitution

: \int \frac{u ( u - 1 )}{512} d u

= \int \frac{u ( u - 1 )}{512} d u

Take the constant out:

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

= \frac{1}{512} \cdot \int u (u - 1) d u

Expand

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{512} \cdot \int u^{\frac{3}{2}} - u 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

= \frac{1}{512} (\int u^{\frac{3}{2}} d u - \int u d u)

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

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

= \frac{1}{512} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + 16 x^{2}

= \frac{1}{512} (\frac{2}{5} (1 + 16 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 16 x^{2})^{\frac{3}{2}})

Simplify

\frac{1}{512} (\frac{2}{5} (1 + 16 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 16 x^{2})^{\frac{3}{2}}) : \frac{1}{1920} (24 x^{2} - 1) (16 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{1920} (24 x^{2} - 1) (16 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{1920} (24 x^{2} - 1) (16 x^{2} + 1)^{\frac{3}{2}} + C

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

\int \frac{3 x + 1}{x ^{3} - 4 x ^{2} + x + 6} d x

\int (t - 1) (2 + t^{2}) d t

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

\int y \cdot cos (y) d y

What is the integral of x^3sqrt(1+16x^2) ?
The integral of x^3sqrt(1+16x^2) is 1/1920 (24x^2-1)(16x^2+1)^{3/2}+C