What is the integral of x\sqrt[5]{(x+1)^2} ?

The integral of x\sqrt[5]{(x+1)^2} is 5/12 (x+1)^{12/5}-5/7 (x+1)^{7/5}+C

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integral of x\sqrt[5]{(x+1)^2}

\frac{5}{12} (x + 1)^{\frac{12}{5}} - \frac{5}{7} (x + 1)^{\frac{7}{5}} + C

Solution steps

assuming

(x + 1) \geq 0

= \int x (x + 1)^{\frac{2}{5}} d x

Apply u-substitution

: \int (u - 1) u^{\frac{2}{5}} d u

= \int (u - 1) u^{\frac{2}{5}} d u

Expand

(u - 1) u^{\frac{2}{5}} : u^{\frac{7}{5}} - u^{\frac{2}{5}}

= \int u^{\frac{7}{5}} - u^{\frac{2}{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

= \int u^{\frac{7}{5}} d u - \int u^{\frac{2}{5}} d u

\int u^{\frac{7}{5}} d u = \frac{5}{12} u^{\frac{12}{5}}

\int u^{\frac{2}{5}} d u = \frac{5}{7} u^{\frac{7}{5}}

= \frac{5}{12} u^{\frac{12}{5}} - \frac{5}{7} u^{\frac{7}{5}}

Substitute back

u = x + 1

= \frac{5}{12} (x + 1)^{\frac{12}{5}} - \frac{5}{7} (x + 1)^{\frac{7}{5}}

Add a constant to the solution

= \frac{5}{12} (x + 1)^{\frac{12}{5}} - \frac{5}{7} (x + 1)^{\frac{7}{5}} + C

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What is the integral of x\sqrt[5]{(x+1)^2} ?
The integral of x\sqrt[5]{(x+1)^2} is 5/12 (x+1)^{12/5}-5/7 (x+1)^{7/5}+C