What is the integral of (16x+3)sqrt(8x^2+3x) ?

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

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integral of (16x+3)sqrt(8x^2+3x)

\int (16 x + 3) 8 x^{2} + 3 x d x

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

Solution steps

\int (16 x + 3) 8 x^{2} + 3 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} + 3 x

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

Add a constant to the solution

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

\frac{d}{d x} ((x - 2) (2 x + 3))

y = \frac{5 x}{1 + x ^{2}}, (4, \frac{20}{17})

\int \frac{7 - 98 x}{81 - 49 x ^{2}} d x

y = x^{0.4}

\frac{d}{d x} ((x + 1)^{3} (x - 2))

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