What is the integral of x^2sqrt(5x+4) ?

The integral of x^2sqrt(5x+4) is 1/125 (2/7 (5x+4)^{7/2}-16/5 (5x+4)^{5/2}+32/3 (5x+4)^{3/2})+C

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integral of x^2sqrt(5x+4)

\int x^{2} 5 x + 4 d x

\frac{1}{125} (\frac{2}{7} (5 x + 4)^{\frac{7}{2}} - \frac{16}{5} (5 x + 4)^{\frac{5}{2}} + \frac{32}{3} (5 x + 4)^{\frac{3}{2}}) + C

Solution steps

\int x^{2} 5 x + 4 d x

Apply u-substitution

: \int \frac{u ( u - 4 ) ^{2}}{125} d u

= \int \frac{u ( u - 4 ) ^{2}}{125} d u

Take the constant out:

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

= \frac{1}{125} \cdot \int u (u - 4)^{2} d u

Expand

u (u - 4)^{2} : u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 u

= \frac{1}{125} \cdot \int u^{\frac{5}{2}} - 8 u^{\frac{3}{2}} + 16 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

= \frac{1}{125} (\int u^{\frac{5}{2}} d u - \int 8 u^{\frac{3}{2}} d u + \int 16 u d u)

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

\int 8 u^{\frac{3}{2}} d u = \frac{16}{5} u^{\frac{5}{2}}

\int 16 u d u = \frac{32}{3} u^{\frac{3}{2}}

= \frac{1}{125} (\frac{2}{7} u^{\frac{7}{2}} - \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}})

Substitute back

u = 5 x + 4

= \frac{1}{125} (\frac{2}{7} (5 x + 4)^{\frac{7}{2}} - \frac{16}{5} (5 x + 4)^{\frac{5}{2}} + \frac{32}{3} (5 x + 4)^{\frac{3}{2}})

Add a constant to the solution

= \frac{1}{125} (\frac{2}{7} (5 x + 4)^{\frac{7}{2}} - \frac{16}{5} (5 x + 4)^{\frac{5}{2}} + \frac{32}{3} (5 x + 4)^{\frac{3}{2}}) + C

f (x) = x^{3} - 6 x^{2} + 8, g (x) = x^{2} - 4 x

\frac{d y}{d x} = (x - 8) e^{- 2 y}

f (x) = 2 x + 6

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

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

What is the integral of x^2sqrt(5x+4) ?
The integral of x^2sqrt(5x+4) is 1/125 (2/7 (5x+4)^{7/2}-16/5 (5x+4)^{5/2}+32/3 (5x+4)^{3/2})+C