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

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

\int (x^{2} 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

x \to 0 + lim (\frac{7 x}{sin ( x )})

f (x) = (x^{4} - 2 x + 8)^{6}, at x = 1

\int e^{x^{2}} 2 x^{3} d x

x \to \infty lim (- 5)

f (t) = arcsin (8 t)

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