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

The integral of x/(sqrt(x-5)) is 2(1/3 (x-5)^{3/2}+5sqrt(x-5))+C

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integral of x/(sqrt(x-5))

\int \frac{x}{x - 5} d x

2 (\frac{1}{3} (x - 5)^{\frac{3}{2}} + 5 x - 5) + C

Solution steps

\int \frac{x}{x - 5} d x

Apply u-substitution

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

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

Take the constant out:

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

= 2 \cdot \int u^{2} + 5 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

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

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

\int 5 d u = 5 u

= 2 (\frac{u ^{3}}{3} + 5 u)

Substitute back

u = x - 5

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

Simplify

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

= 2 (\frac{1}{3} (x - 5)^{\frac{3}{2}} + 5 x - 5)

Add a constant to the solution

= 2 (\frac{1}{3} (x - 5)^{\frac{3}{2}} + 5 x - 5) + C

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

f (x) = (\frac{x x - 1}{2})

m (r) = \frac{r ^{3} + 3 r + 2}{r ^{2} - 1}

\frac{d}{d x} (5 x - 1 - 2 x)

- 5 x^{\frac{1}{2}} + x^{\frac{3}{2}}

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