What is the integral of (6x+1)sqrt(3x^2+x) ?

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

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integral of (6x+1)sqrt(3x^2+x)

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

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

Solution steps

\int (6 x + 1) 3 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 = 3 x^{2} + x

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (\frac{1}{x y ^{2} + y x ^{2}})

\int \frac{( 2 + ln ( 7 x ) )}{3 x} d x

6 x^{2} - 48 x + y^{2} + 5 = 0

x \to (- k + 1) lim (\frac{1}{x - 2})

y^{^{'}} + 8 y = 0

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