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

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

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integral of x^2sqrt(x-4)

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

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

Solution steps

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

Apply u-substitution

: \int (u + 4)^{2} u d u

= \int (u + 4)^{2} u d u

Expand

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

= \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

= \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{2}{7} u^{\frac{7}{2}} + \frac{16}{5} u^{\frac{5}{2}} + \frac{32}{3} u^{\frac{3}{2}}

Substitute back

u = x - 4

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

Add a constant to the solution

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

\int \frac{sin ^{3} ( x )}{cos ^{5} ( x )} d x

\frac{\partial}{\partial y} (e^{3 x - y^{2}})

\frac{d ^{2}}{d x ^{2}} (\frac{x}{e ^{x}})

\frac{d}{d x} (arccos (e^{5 x}))

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