What is the integral of (6x+5)sqrt(3x^2+5x) ?

The integral of (6x+5)sqrt(3x^2+5x) is 2/3 (3x^2+5x)^{3/2}+C

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integral of (6x+5)sqrt(3x^2+5x)

\int (6 x + 5) 3 x^{2} + 5 x d x

\frac{2}{3} (3 x^{2} + 5 x)^{\frac{3}{2}} + C

Solution steps

\int (6 x + 5) 3 x^{2} + 5 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 = 3 x^{2} + 5 x

= \frac{2}{3} (3 x^{2} + 5 x)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{3} (3 x^{2} + 5 x)^{\frac{3}{2}} + C

t \to 0 lim (\frac{4 t ^{2} + 3 t + 2}{t ^{3} + 2 t - 6})

f (x) = - x^{2} + 3 x - 1

\int \frac{( e ^{2 x} )}{e ^{4 x} + 4} d x

\int \frac{1}{8 - 2 x - x ^{2}} d x

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

What is the integral of (6x+5)sqrt(3x^2+5x) ?
The integral of (6x+5)sqrt(3x^2+5x) is 2/3 (3x^2+5x)^{3/2}+C