What is the integral of x^3sqrt(1+9x^4) ?

The integral of x^3sqrt(1+9x^4) is 1/54 (1+9x^4)^{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+9x^4)

\int x^{3} 1 + 9 x^{4} d x

\frac{1}{54} (1 + 9 x^{4})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 1 + 9 x^{4} d x

Apply u-substitution

: \int \frac{u}{36} d u

= \int \frac{u}{36} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{36} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 9 x^{4}

= \frac{1}{36} \cdot \frac{2}{3} (1 + 9 x^{4})^{\frac{3}{2}}

Simplify

\frac{1}{36} \cdot \frac{2}{3} (1 + 9 x^{4})^{\frac{3}{2}} : \frac{1}{54} (1 + 9 x^{4})^{\frac{3}{2}}

= \frac{1}{54} (1 + 9 x^{4})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{54} (1 + 9 x^{4})^{\frac{3}{2}} + C

\frac{- 3 s - 4}{2 ( s ^{2} + 2 s + 2 )}

\frac{e ^{- 8 s}}{s}

\frac{d}{d x} (e^{2 e^{x}})

\int ((csc (y))^{2} - sec (y) tan (y)) d y

\int_{- π}^{π} x^{2} sin (x) d x

What is the integral of x^3sqrt(1+9x^4) ?
The integral of x^3sqrt(1+9x^4) is 1/54 (1+9x^4)^{3/2}+C