What is the integral of (4x+4)sqrt(2x^2+4x) ?

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

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integral of (4x+4)sqrt(2x^2+4x)

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

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

Solution steps

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

(4 x + 4) 2 x^{2} + 4 x = 42 (x + 1) x^{2} + 2 x

= \int 42 (x + 1) x^{2} + 2 x d x

Take the constant out:

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

= 42 \cdot \int (x + 1) x^{2} + 2 x d x

Apply u-substitution

: \int u^{2} d u

= 42 \cdot \int u^{2} d u

Apply the Power Rule

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

= 42 \frac{u ^{3}}{3}

Substitute back

u = x^{2} + 2 x

= 42 \frac{( x ^{2} + 2 x ) ^{3}}{3}

Simplify

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

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

Add a constant to the solution

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

\int 50 x d x

x - 1, x - 1

\int_{0}^{π} 3 e^{x} sin (x) d x

\frac{d}{d x} (\frac{1}{9 - x ^{2}})

\frac{1}{( s + 1 ) ^{2} \cdot ( s - 1 )}

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