What is the integral of ((x+3))/((x^2+6x)^{1/3)} ?

The integral of ((x+3))/((x^2+6x)^{1/3)} is 3/4 (x^2+6x)^{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+3))/((x^2+6x)^{1/3)}

\int \frac{( x + 3 )}{( x ^{2} + 6 x ) ^{\frac{1}{3}}} d x

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

Solution steps

\int \frac{x + 3}{( x ^{2} + 6 x ) ^{\frac{1}{3}}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{1}{3}}} d u

= \int \frac{1}{2 u ^{\frac{1}{3}}} 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 \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Power Rule

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

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

Substitute back

u = x^{2} + 6 x

= \frac{1}{2} \cdot \frac{3}{2} (x^{2} + 6 x)^{\frac{2}{3}}

Simplify

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

= \frac{3}{4} (x^{2} + 6 x)^{\frac{2}{3}}

Add a constant to the solution

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

\frac{d y}{d x} = \frac{3 x ^{4} + y ^{3}}{3 x y ^{2}}

\frac{d y}{d t} = \frac{1}{y - 2}

\int_{0}^{1} 4 x y e^{- (2 x + y)} d y

n = 0 \sum \infty (- \frac{5}{6})^{n}

\int cos^{2} (θ) + sin^{2} (θ) d θ

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