What is the integral of x/(sqrt(x-10)) ?

The integral of x/(sqrt(x-10)) is 2(1/3 (x-10)^{3/2}+10sqrt(x-10))+C

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integral of x/(sqrt(x-10))

\int \frac{x}{x - 10} d x

2 (\frac{1}{3} (x - 10)^{\frac{3}{2}} + 10 x - 10) + C

Solution steps

\int \frac{x}{x - 10} d x

Apply u-substitution

: \int 2 (u^{2} + 10) d u

= \int 2 (u^{2} + 10) 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} + 10 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 10 d u)

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

\int 10 d u = 10 u

= 2 (\frac{u ^{3}}{3} + 10 u)

Substitute back

u = x - 10

= 2 (\frac{( x - 10 ) ^{3}}{3} + 10 x - 10)

Simplify

2 (\frac{( x - 10 ) ^{3}}{3} + 10 x - 10) : 2 (\frac{1}{3} (x - 10)^{\frac{3}{2}} + 10 x - 10)

= 2 (\frac{1}{3} (x - 10)^{\frac{3}{2}} + 10 x - 10)

Add a constant to the solution

= 2 (\frac{1}{3} (x - 10)^{\frac{3}{2}} + 10 x - 10) + C

n = 1 \sum \infty n 2

y^{^{'}} - 2 = 2 y^{2}

(4, 0), (0, - 1)

\frac{d}{d x} (\frac{5 + 6 x}{9 - 4 x})

f (1) = x + \frac{1}{x}

What is the integral of x/(sqrt(x-10)) ?
The integral of x/(sqrt(x-10)) is 2(1/3 (x-10)^{3/2}+10sqrt(x-10))+C