What is the integral of x/(sqrt(x-8)) ?

The integral of x/(sqrt(x-8)) is 2(1/3 (x-8)^{3/2}+8sqrt(x-8))+C

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integral of x/(sqrt(x-8))

\int \frac{x}{x - 8} d x

2 (\frac{1}{3} (x - 8)^{\frac{3}{2}} + 8 x - 8) + C

Solution steps

\int \frac{x}{x - 8} d x

Apply u-substitution

: \int 2 (u^{2} + 8) d u

= \int 2 (u^{2} + 8) 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} + 8 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 8 d u)

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

\int 8 d u = 8 u

= 2 (\frac{u ^{3}}{3} + 8 u)

Substitute back

u = x - 8

= 2 (\frac{( x - 8 ) ^{3}}{3} + 8 x - 8)

Simplify

2 (\frac{( x - 8 ) ^{3}}{3} + 8 x - 8) : 2 (\frac{1}{3} (x - 8)^{\frac{3}{2}} + 8 x - 8)

= 2 (\frac{1}{3} (x - 8)^{\frac{3}{2}} + 8 x - 8)

Add a constant to the solution

= 2 (\frac{1}{3} (x - 8)^{\frac{3}{2}} + 8 x - 8) + C

\int_{0}^{1} 2 x arctan (x^{2}) d x

\frac{d y}{d x} = 4 x^{5} y - y

\int_{- 2}^{0} (\frac{y}{2} + y + 3) d y

e^{- 2 t} sin (5 t)

6 e^{x^{- 5}}

What is the integral of x/(sqrt(x-8)) ?
The integral of x/(sqrt(x-8)) is 2(1/3 (x-8)^{3/2}+8sqrt(x-8))+C