What is the integral of (16x+1)sqrt(8x^2+x) ?

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

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integral of (16x+1)sqrt(8x^2+x)

\int (16 x + 1) 8 x^{2} + x d x

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

Solution steps

\int (16 x + 1) 8 x^{2} + 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} + x

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

Add a constant to the solution

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

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