What is the integral of 1/((a+bx)^{1/3)} ?

The integral of 1/((a+bx)^{1/3)} is (3(a+bx)^{2/3})/(2b)+C

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integral of 1/((a+bx)^{1/3)}

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

\frac{3 ( a + b x ) ^{\frac{2}{3}}}{2 b} + C

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{b 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}{b} \cdot \int \frac{1}{u ^{\frac{1}{3}}} d u

Apply the Power Rule

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

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

Substitute back

u = a + b x

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

Simplify

\frac{1}{b} \cdot \frac{3}{2} (a + b x)^{\frac{2}{3}} : \frac{3 ( a + b x ) ^{\frac{2}{3}}}{2 b}

= \frac{3 ( a + b x ) ^{\frac{2}{3}}}{2 b}

Add a constant to the solution

= \frac{3 ( a + b x ) ^{\frac{2}{3}}}{2 b} + C

x^{2} y d x + x^{3} d y = 0

x \to - 2 lim (- 2 x^{3} - 3 x^{2} + 8)

\int 0.1 e^{0.4 x} d x

y^{^{'''}} + 5 y^{^{''}} + 7 y^{^{'}} - 13 y = 0

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

What is the integral of 1/((a+bx)^{1/3)} ?
The integral of 1/((a+bx)^{1/3)} is (3(a+bx)^{2/3})/(2b)+C