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

The integral of x^4sqrt(x^5-4) is 2/15 (x^5-4)^{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^4sqrt(x^5-4)

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

\frac{2}{15} (x^{5} - 4)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int \frac{u}{5} d u

= \int \frac{u}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{5} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = x^{5} - 4

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

Simplify

\frac{1}{5} \cdot \frac{2}{3} (x^{5} - 4)^{\frac{3}{2}} : \frac{2}{15} (x^{5} - 4)^{\frac{3}{2}}

= \frac{2}{15} (x^{5} - 4)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{15} (x^{5} - 4)^{\frac{3}{2}} + C

\int 4 cos^{2} (2 x) d x

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

\int (x^{\frac{5}{2}} + 2 x^{\frac{3}{2}} - x) d x

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

What is the integral of x^4sqrt(x^5-4) ?
The integral of x^4sqrt(x^5-4) is 2/15 (x^5-4)^{3/2}+C