What is the integral of x/(\sqrt[5]{2x^2-4)} ?

The integral of x/(\sqrt[5]{2x^2-4)} is 5/16 (2x^2-4)^{4/5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x/(\sqrt[5]{2x^2-4)}

\int \frac{x}{5 2 x ^{2} - 4} d x

\frac{5}{16} (2 x^{2} - 4)^{\frac{4}{5}} + C

Solution steps

\int \frac{x}{5 2 x ^{2} - 4} d x

Apply u-substitution

= \int \frac{1}{4 5 u} 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 \frac{1}{5 u} d u

Apply the Power Rule

= \frac{1}{4} \cdot \frac{5}{4} u^{\frac{4}{5}}

Substitute back

u = 2 x^{2} - 4

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

Simplify

\frac{1}{4} \cdot \frac{5}{4} (2 x^{2} - 4)^{\frac{4}{5}} : \frac{5}{16} (2 x^{2} - 4)^{\frac{4}{5}}

= \frac{5}{16} (2 x^{2} - 4)^{\frac{4}{5}}

Add a constant to the solution

= \frac{5}{16} (2 x^{2} - 4)^{\frac{4}{5}} + C

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

\int 3 cos (t) sin (t) d t

\int \frac{d}{1} (8 x) d x

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

\int sin (π x) sin (2 π x) d x

What is the integral of x/(\sqrt[5]{2x^2-4)} ?
The integral of x/(\sqrt[5]{2x^2-4)} is 5/16 (2x^2-4)^{4/5}+C