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

The integral of x^2sqrt(5x+5) is sqrt(5)(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(5x+5)

\int x^{2} 5 x + 5 d x

5 (\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} 5 x + 5 d x

x^{2} 5 x + 5 = 5 x^{2} x + 1

= \int 5 x^{2} x + 1 d x

Take the constant out:

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

= 5 \cdot \int x^{2} x + 1 d x

Apply u-substitution

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

= 5 \cdot \int (u - 1)^{2} u d u

Expand

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

= 5 \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

= 5 (\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}}

= 5 (\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

= 5 (\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

= 5 (\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

\int \frac{5}{33 x} x^{4} d x

t \to 0 lim (- 2 t)

\int \frac{- 7 25 x ^{2} - 9}{x ^{3}} d x

x \to 0 lim ((1 + 2 x)^{\frac{3}{x}})

y^{^{''}} - 7 y^{^{'}} + 14 y = - 784 t^{2}

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