Integral Calculus: Evaluate the integral of 1/(5x-sqrt(x))?

Question

Please give full working out if that is not too much trouble. Or at least the key steps would be nice! Have fun. This is an antidifferentitiation question.

Answer 1

∫1/(5x - √x) dx

= ∫1/(√x(5√x - 1)) dx

let u = 5√x - 1

du = 5/(2√x) dx

= ∫1/(√x(5√x - 1)) dx

= 2/5*∫5/(2√x(5√x - 1)) dx

= 2/5*∫du/u

= 2/5 * ln | u | + C

= 2/5 * ln | 5√x - 1 | + C

Answer 2

∫ 1 / [ 5·x - √(x) ] dx

Let u = √(x)

Then du = (x)^(-½)·(½) dx = dx/[ 2·√(x) ]

And dx = 2·√(x) du = 2·u du

→ ∫ 1 / [ 5·u² - u ] 2·u du

= ∫ 1 / { u·[ 5·u - 1 ] } 2·u du

= ∫ 2 / [ 5·u - 1 ] du

Let v = 5·u - 1

Then dv = 5 du

And du = dv/5

→ ∫ 2 / [ v ] dv/5

= (2/5) · ∫ 1 / v dv

Integrate:

= (2/5) · ln| v | + C

Reverse substitution for v and then u:

= (2/5) · ln| 5·u - 1 | + C

= (2/5) · ln| 5·√(x) - 1 | + C

Answer:

∫ 1 / [ 5·x - √(x) ] dx = (2/5) · ln| 5·√(x) - 1 | + C

Answer 3

Use substitution t^2 = x. So dx = 2tdt.

So integral = S tdt/(5t^2 - t)...................S is integral

                = S 2dt/(5t-1)

                 = (2/5)ln|5t - 1| + c.

                  = (2/5) ln|5rtx - 1| +c.