What is the integral of (12x+2)sqrt(6x^2+2x) ?

The integral of (12x+2)sqrt(6x^2+2x) is (4sqrt(2))/3 (3x^2+x)^{3/2}+C

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integral of (12x+2)sqrt(6x^2+2x)

\int (12 x + 2) 6 x^{2} + 2 x d x

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

Solution steps

\int (12 x + 2) 6 x^{2} + 2 x d x

(12 x + 2) 6 x^{2} + 2 x = 22 (6 x + 1) 3 x^{2} + x

= \int 22 (6 x + 1) 3 x^{2} + x d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 22 \cdot \int (6 x + 1) 3 x^{2} + x d x

Apply u-substitution

: \int u d u

= 22 \cdot \int u d u

Apply the Power Rule

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

= 22 \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x^{2} + x

= 22 \frac{2}{3} (3 x^{2} + x)^{\frac{3}{2}}

Simplify

22 \frac{2}{3} (3 x^{2} + x)^{\frac{3}{2}} : \frac{4 2}{3} (3 x^{2} + x)^{\frac{3}{2}}

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

Add a constant to the solution

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

\int x e^{- 3 x^{2}} d x

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\int \frac{1}{a + b x ^{2}} d x

\int (sin (x))^{2} cos (x) d x

x \to 0 lim (\frac{0}{2 x ^{2}})

What is the integral of (12x+2)sqrt(6x^2+2x) ?
The integral of (12x+2)sqrt(6x^2+2x) is (4sqrt(2))/3 (3x^2+x)^{3/2}+C