What is the integral of x^3sqrt(5-x^4) ?

The integral of x^3sqrt(5-x^4) is -1/6 (5-x^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(5-x^4)

\int x^{3} 5 - x^{4} d x

- \frac{1}{6} (5 - x^{4})^{\frac{3}{2}} + C

Solution steps

\int x^{3} 5 - x^{4} d x

Apply u-substitution

: \int - \frac{u}{4} d u

= \int - \frac{u}{4} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= - \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 5 - x^{4}

= - \frac{1}{4} \cdot \frac{2}{3} (5 - x^{4})^{\frac{3}{2}}

Simplify

- \frac{1}{4} \cdot \frac{2}{3} (5 - x^{4})^{\frac{3}{2}} : - \frac{1}{6} (5 - x^{4})^{\frac{3}{2}}

= - \frac{1}{6} (5 - x^{4})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (5 - x^{4})^{\frac{3}{2}} + C

\int (4 x + 29)^{- 2} d x

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

\int \frac{1}{x + 1 - x} d x

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

\int \frac{( sin ^{2} ( \frac{1}{t} ) )}{t ^{2}} d t

What is the integral of x^3sqrt(5-x^4) ?
The integral of x^3sqrt(5-x^4) is -1/6 (5-x^4)^{3/2}+C