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

The integral of 1/(x^3(2+x^3)^{1/3)} is -1/(2^{1+1/3)x^2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^3(2+x^3)^{1/3)}

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

- \frac{1}{2 ^{1 + \frac{1}{3}} x ^{2}} + C

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

= \int x^{- 3} \frac{1}{2 ^{\frac{1}{3}}} d x

Take the constant out:

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

= \frac{1}{2 ^{\frac{1}{3}}} \cdot \int x^{- 3} d x

Apply the Power Rule

: - \frac{1}{2 x ^{2}}

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

Simplify

\frac{1}{2 ^{\frac{1}{3}}} (- \frac{1}{2 x ^{2}}) : - \frac{1}{2 ^{1 + \frac{1}{3}} x ^{2}}

= - \frac{1}{2 ^{1 + \frac{1}{3}} x ^{2}}

Add a constant to the solution

= - \frac{1}{2 ^{1 + \frac{1}{3}} x ^{2}} + C

\int \frac{4 x ^{3} - 2 x ^{2} + 5}{x ^{2} - x - 6} d x

\int 10 (20 - x) e^{- \frac{x}{20}} d x

\int t^{3} - \frac{e ^{- t} - 4}{e ^{- t}} d t

\int (tan (x))^{\frac{1}{3}} sec^{4} (x) d x

\int 3 t 2 - 2 t^{2} d t

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