What is the integral of (16x+4)sqrt(8x^2+4x) ?

The integral of (16x+4)sqrt(8x^2+4x) is 16/3 (x(2x+1))^{3/2}+C

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integral of (16x+4)sqrt(8x^2+4x)

\int (16 x + 4) 8 x^{2} + 4 x d x

\frac{16}{3} (x (2 x + 1))^{\frac{3}{2}} + C

Solution steps

\int (16 x + 4) 8 x^{2} + 4 x d x

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 8 x^{2} + 4 x

= \frac{2}{3} (8 x^{2} + 4 x)^{\frac{3}{2}}

Simplify

\frac{2}{3} (8 x^{2} + 4 x)^{\frac{3}{2}} : \frac{16}{3} (x (2 x + 1))^{\frac{3}{2}}

= \frac{16}{3} (x (2 x + 1))^{\frac{3}{2}}

Add a constant to the solution

= \frac{16}{3} (x (2 x + 1))^{\frac{3}{2}} + C

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What is the integral of (16x+4)sqrt(8x^2+4x) ?
The integral of (16x+4)sqrt(8x^2+4x) is 16/3 (x(2x+1))^{3/2}+C