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

The integral of x^3sqrt(1+3x^2) is 1/135 (9x^2-2)(3x^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+3x^2)

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

\frac{1}{135} (9 x^{2} - 2) (3 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( u - 1 )}{18} d u

= \int \frac{u ( u - 1 )}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u (u - 1) d u

Expand

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

= \frac{1}{18} \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}{18} (\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}{18} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 + 3 x^{2}

= \frac{1}{18} (\frac{2}{5} (1 + 3 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 3 x^{2})^{\frac{3}{2}})

Simplify

\frac{1}{18} (\frac{2}{5} (1 + 3 x^{2})^{\frac{5}{2}} - \frac{2}{3} (1 + 3 x^{2})^{\frac{3}{2}}) : \frac{1}{135} (9 x^{2} - 2) (3 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{135} (9 x^{2} - 2) (3 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{135} (9 x^{2} - 2) (3 x^{2} + 1)^{\frac{3}{2}} + C

\int \frac{x ^{2} - 6}{x ^{2} - 9} d x

\int 20 - \frac{t}{5} d t

\int t^{\frac{7}{2}} (t^{2} - 10 t + 17) d t

\int \frac{sec ^{2} ( x )}{1 - tan ( x )} d x

\int t e^{\frac{1}{4} l n (t)} d t

What is the integral of x^3sqrt(1+3x^2) ?
The integral of x^3sqrt(1+3x^2) is 1/135 (9x^2-2)(3x^2+1)^{3/2}+C