What is the integral of x\sqrt[5]{4+x^2} ?

The integral of x\sqrt[5]{4+x^2} is 5/12 (4+x^2)^{6/5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x\sqrt[5]{4+x^2}

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

\frac{5}{12} (4 + x^{2})^{\frac{6}{5}} + C

Solution steps

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

Apply u-substitution

: \int \frac{5 u}{2} d u

= \int \frac{5 u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 5 u d u

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{1}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = 4 + x^{2}

= \frac{1}{2} \cdot \frac{5}{6} (4 + x^{2})^{\frac{6}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{6} (4 + x^{2})^{\frac{6}{5}} : \frac{5}{12} (4 + x^{2})^{\frac{6}{5}}

= \frac{5}{12} (4 + x^{2})^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{12} (4 + x^{2})^{\frac{6}{5}} + C

\int 8 cos^{2} (θ) d θ

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

\int sec (w) d w

\int \frac{1}{( 1 + 3 x )} d x

\int 4 x I n (8 x + 6)

What is the integral of x\sqrt[5]{4+x^2} ?
The integral of x\sqrt[5]{4+x^2} is 5/12 (4+x^2)^{6/5}+C