What is the integral of 1/(\sqrt[3]{(x-1)^2)} ?

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

Popular >

integral of 1/(\sqrt[3]{(x-1)^2)}

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

3 3 x - 1 + C

Solution steps

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

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

assuming

(x - 1) \geq 0

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

Apply u-substitution

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

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

Apply the Power Rule

: 3 u^{\frac{1}{3}}

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

Substitute back

u = x - 1

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

Apply radical rule:

a^{\frac{1}{n}} = n a

= 3 3 x - 1

Add a constant to the solution

= 3 3 x - 1 + C

\frac{d}{d x} (\frac{1}{5 x ^{3}})

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

7 cos (2 x)

x \to 0 + lim ((1 - ln (x))^{x})

\int_{0}^{a} a^{2} + x^{2} d x

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