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

The integral of x^2*sqrt(x-5) is 2/7 (x-5)^{7/2}+4(x-5)^{5/2}+50/3 (x-5)^{3/2}+C

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integral of x^2*sqrt(x-5)

\int x^{2} \cdot x - 5 d x

\frac{2}{7} (x - 5)^{\frac{7}{2}} + 4 (x - 5)^{\frac{5}{2}} + \frac{50}{3} (x - 5)^{\frac{3}{2}} + C

Solution steps

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

Apply u-substitution

: \int (u + 5)^{2} u d u

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

Expand

(u + 5)^{2} u : u^{\frac{5}{2}} + 10 u^{\frac{3}{2}} + 25 u

= \int u^{\frac{5}{2}} + 10 u^{\frac{3}{2}} + 25 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 10 u^{\frac{3}{2}} d u + \int 25 u d u

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

\int 10 u^{\frac{3}{2}} d u = 4 u^{\frac{5}{2}}

\int 25 u d u = \frac{50}{3} u^{\frac{3}{2}}

= \frac{2}{7} u^{\frac{7}{2}} + 4 u^{\frac{5}{2}} + \frac{50}{3} u^{\frac{3}{2}}

Substitute back

u = x - 5

= \frac{2}{7} (x - 5)^{\frac{7}{2}} + 4 (x - 5)^{\frac{5}{2}} + \frac{50}{3} (x - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{7} (x - 5)^{\frac{7}{2}} + 4 (x - 5)^{\frac{5}{2}} + \frac{50}{3} (x - 5)^{\frac{3}{2}} + C

y = x^{2} (x^{3} - 1)^{5}

y = 2 x + 1, at x = 4

(x - y - 1)^{3} = x

n = 1 \sum \infty \frac{x ^{n}}{3 ^{n}}

\frac{d}{d x} (\frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9})

What is the integral of x^2*sqrt(x-5) ?
The integral of x^2*sqrt(x-5) is 2/7 (x-5)^{7/2}+4(x-5)^{5/2}+50/3 (x-5)^{3/2}+C