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

The integral of x^2sqrt(4x+4) is 2(2/7 (x+1)^{7/2}-4/5 (x+1)^{5/2}+2/3 (x+1)^{3/2})+C

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

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

2 (\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}}) + C

Solution steps

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

x^{2} 4 x + 4 = 4 x^{2} x + 1

= \int 4 x^{2} x + 1 d x

Take the constant out:

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

= 4 \cdot \int x^{2} x + 1 d x

4 = 2

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

Apply u-substitution

= 2 \cdot \int (u - 1)^{2} u d u

Expand

(u - 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

= 2 \cdot \int u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int u^{\frac{5}{2}} d u - \int 2 u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

\int 2 u^{\frac{3}{2}} d u = \frac{4}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= 2 (\frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = x + 1

= 2 (\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}})

Add a constant to the solution

= 2 (\frac{2}{7} (x + 1)^{\frac{7}{2}} - \frac{4}{5} (x + 1)^{\frac{5}{2}} + \frac{2}{3} (x + 1)^{\frac{3}{2}}) + C

d er i v a t i v e cos (3 t) sin (2 t)

\int \frac{6}{sin ^{2} ( x )} d x

d er i v a t i v e f (x) = \frac{x ^{3}}{3} + x^{2} + 4 x - 10

t an g e n t y = 10 x - x^{2}, (5, 26)

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

What is the integral of x^2sqrt(4x+4) ?
The integral of x^2sqrt(4x+4) is 2(2/7 (x+1)^{7/2}-4/5 (x+1)^{5/2}+2/3 (x+1)^{3/2})+C