What is the integral of x\sqrt[5]{x-1} ?

The integral of x\sqrt[5]{x-1} is 5(1/11 (x-1)^{11/5}+1/6 (x-1)^{6/5})+C

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integral of x\sqrt[5]{x-1}

5 (\frac{1}{11} (x - 1)^{\frac{11}{5}} + \frac{1}{6} (x - 1)^{\frac{6}{5}}) + C

Solution steps

Apply u-substitution

: \int 5 u^{5} (u^{5} + 1) d u

= \int 5 u^{5} (u^{5} + 1) d u

Take the constant out:

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

= 5 \cdot \int u^{5} (u^{5} + 1) d u

Expand

u^{5} (u^{5} + 1) : u^{10} + u^{5}

= 5 \cdot \int u^{10} + u^{5} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (\int u^{10} d u + \int u^{5} d u)

\int u^{10} d u = \frac{u ^{11}}{11}

\int u^{5} d u = \frac{u ^{6}}{6}

= 5 (\frac{u ^{11}}{11} + \frac{u ^{6}}{6})

Substitute back

Simplify

= 5 (\frac{1}{11} (x - 1)^{\frac{11}{5}} + \frac{1}{6} (x - 1)^{\frac{6}{5}})

Add a constant to the solution

= 5 (\frac{1}{11} (x - 1)^{\frac{11}{5}} + \frac{1}{6} (x - 1)^{\frac{6}{5}}) + C

\int_{- 2}^{3} \frac{39}{x ^{4}} d x

f (θ) = cos (θ^{2})

\frac{\partial}{\partial x} (ln (y^{2}))

y^{^{''}} + b y (x) = 0

g (x) = x^{2} + 3 x

What is the integral of x\sqrt[5]{x-1} ?
The integral of x\sqrt[5]{x-1} is 5(1/11 (x-1)^{11/5}+1/6 (x-1)^{6/5})+C