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

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

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integral of x/(sqrt(2x-5))

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2} + 5}{2} d u

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

Take the constant out:

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

= \frac{1}{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

= \frac{1}{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

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

Substitute back

u = 2 x - 5

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{1}{x} (1.1)

\int \frac{x}{x ^{2} - 4 x + 4} d x

\int \frac{- 8 x - 9}{x ^{9}} d x

4 x^{3} - 3 x^{2} + 4, at x = 2

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