What is the integral of (x^4)/(\sqrt[3]{9+x^5)} ?

The integral of (x^4)/(\sqrt[3]{9+x^5)} is 3/10 (9+x^5)^{2/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^4)/(\sqrt[3]{9+x^5)}

\int \frac{x ^{4}}{3 9 + x ^{5}} d x

\frac{3}{10} (9 + x^{5})^{\frac{2}{3}} + C

Solution steps

\int \frac{x ^{4}}{3 9 + x ^{5}} d x

Apply u-substitution

: \int \frac{1}{5 3 u} d u

= \int \frac{1}{5 3 u} 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 \frac{1}{3 u} d u

Apply the Power Rule

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

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

Substitute back

u = 9 + x^{5}

= \frac{1}{5} \cdot \frac{3}{2} (9 + x^{5})^{\frac{2}{3}}

Simplify

\frac{1}{5} \cdot \frac{3}{2} (9 + x^{5})^{\frac{2}{3}} : \frac{3}{10} (9 + x^{5})^{\frac{2}{3}}

= \frac{3}{10} (9 + x^{5})^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{10} (9 + x^{5})^{\frac{2}{3}} + C

\int a sin (x) cos (x) d x

\int 3 cos (x) sin^{2} (x) d x

\int \frac{1}{( 4 x ^{2} - 9 ) ^{3}} d x

\int - t cos (2 t) d t

\int 2 x^{3} - 5 x^{2} - 3 x + 4 d x

What is the integral of (x^4)/(\sqrt[3]{9+x^5)} ?
The integral of (x^4)/(\sqrt[3]{9+x^5)} is 3/10 (9+x^5)^{2/3}+C