What is the integral of (6x+1)(6x^2+2x)^{3/4} ?

The integral of (6x+1)(6x^2+2x)^{3/4} is 2/7 (6x^2+2x)^{7/4}+C

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integral of (6x+1)(6x^2+2x)^{3/4}

\int (6 x + 1) (6 x^{2} + 2 x)^{\frac{3}{4}} d x

\frac{2}{7} (6 x^{2} + 2 x)^{\frac{7}{4}} + C

Solution steps

\int (6 x + 1) (6 x^{2} + 2 x)^{\frac{3}{4}} d x

Apply u-substitution

: \int \frac{u ^{\frac{3}{4}}}{2} d u

= \int \frac{u ^{\frac{3}{4}}}{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{3}{4}} d u

Apply the Power Rule

: \frac{4}{7} u^{\frac{7}{4}}

= \frac{1}{2} \cdot \frac{4}{7} u^{\frac{7}{4}}

Substitute back

u = 6 x^{2} + 2 x

= \frac{1}{2} \cdot \frac{4}{7} (6 x^{2} + 2 x)^{\frac{7}{4}}

Simplify

\frac{1}{2} \cdot \frac{4}{7} (6 x^{2} + 2 x)^{\frac{7}{4}} : \frac{2}{7} (6 x^{2} + 2 x)^{\frac{7}{4}}

= \frac{2}{7} (6 x^{2} + 2 x)^{\frac{7}{4}}

Add a constant to the solution

= \frac{2}{7} (6 x^{2} + 2 x)^{\frac{7}{4}} + C

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What is the integral of (6x+1)(6x^2+2x)^{3/4} ?
The integral of (6x+1)(6x^2+2x)^{3/4} is 2/7 (6x^2+2x)^{7/4}+C