Question

How do you integral `int sqrtx/(sqrtx-3)dx`?

Answer 1

`x+6sqrt(x)+18ln(sqrt(x)-3)+C_1`

Explanation:

We want to solve

`intsqrt(x)/(sqrt(x)-3)dx`

Use integration by substitution

Let `u=sqrt(x)-3` then `(du)/dx=1/(2sqrt(x))`

`intsqrt(x)/u*1/(1/(2sqrt(x)))du`

`2int(sqrt(x)sqrt(x))/udu`

But `u=sqrt(x)-3<=>sqrt(x)=u+3`

`2int((u+3)(u+3))/udu`

`2int(u^2+6u+9)/udu`

`2intudu+12intdu+18int1/udu`

Integrate

`u^2+12u+18ln(u)+C`

Substitute `u=sqrt(x)-3`

`(sqrt(x)-3)^2+12(sqrt(x)-3)+18ln(sqrt(x)-3)+C`

Which can be simplified as

`x+6sqrt(x)+18ln(sqrt(x)-3)+C_1`