What is the integral of (x+2)(x^2+4x+10)^{1/5} ?

The integral of (x+2)(x^2+4x+10)^{1/5} is 5/12 (x^2+4x+10)^{6/5}+C

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integral of (x+2)(x^2+4x+10)^{1/5}

\int (x + 2) (x^{2} + 4 x + 10)^{\frac{1}{5}} d x

\frac{5}{12} (x^{2} + 4 x + 10)^{\frac{6}{5}} + C

Solution steps

\int (x + 2) (x^{2} + 4 x + 10)^{\frac{1}{5}} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{\frac{1}{5}} d u

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{1}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = x^{2} + 4 x + 10

= \frac{1}{2} \cdot \frac{5}{6} (x^{2} + 4 x + 10)^{\frac{6}{5}}

Simplify

\frac{1}{2} \cdot \frac{5}{6} (x^{2} + 4 x + 10)^{\frac{6}{5}} : \frac{5}{12} (x^{2} + 4 x + 10)^{\frac{6}{5}}

= \frac{5}{12} (x^{2} + 4 x + 10)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{12} (x^{2} + 4 x + 10)^{\frac{6}{5}} + C

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What is the integral of (x+2)(x^2+4x+10)^{1/5} ?
The integral of (x+2)(x^2+4x+10)^{1/5} is 5/12 (x^2+4x+10)^{6/5}+C