Question

How do you find the indefinite integral of `int sqrtx/(sqrtx-3)`?

Answer 1

I got:

`x + 6sqrtx + 18ln|sqrtx - 3| + C`

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Let `u = sqrtx`. Then, `du = 1/(2sqrtx)dx`, `2udu = dx`, and we have:

`=> 2int u^2/(u - 3)du`

By subtracting and adding `9/(u - 3)`, we take advantage of the difference of two squares, and obtain:

`=> 2int (u^2 - 9)/(u - 3) + 9/(u - 3)du`

`= 2int ((u + 3)cancel((u - 3)))/cancel(u - 3) + 9/(u - 3)du`

`= 2int u+3 + 9/(u - 3)du`

`= 2(u^2/2 + 3u + 9ln|u - 3|)`

`= u^2 + 6u + 18ln|u - 3|`

But since `u = sqrtx`, we un-substitute to get:

`=> color(blue)(int sqrtx/(sqrt(x) - 3)dx = x + 6sqrtx + 18ln|sqrtx - 3| + C)`

Answer 2

`x+6sqrtx+18lnabs(sqrtx-3)+C`

Explanation:

`I=intsqrtx/(sqrtx-3)dx`

We will use the substitution `u=sqrtx-3`. This also implies that `sqrtx=u+3`, which is a substitution we will use in the numerator.

Differentiating both sides of the substitution, we see that `du=1/(2sqrtx)dx`, which looks (and is) a little messy. However, we can modify the integral to force this `1/(2sqrtx)dx` term to appear to our benefit!

Since there's currently no `1/(2sqrtx)` term present, we can multiply the integrand by `2sqrtx1/(2sqrtx)` to have no net change.

`I=intsqrtx/(sqrtx-3)(2sqrtx)(1/(2sqrtx)dx)`

`I=2int(sqrtxsqrtx)/(sqrtx-3)(1/(2sqrtx)dx)`

Returning to our substitutions, which we can now make more easily: we have let `sqrtx-3=u`, and following from this `sqrtx=u+3` and `1/(2sqrtx)dx=du`.

`I=2int(u+3)^2/udu`

Expanding the squared term:

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

Dividing through:

`I=2int(u+6+9/u)du`

We can write this as `3` separate integrals (don't forget to distribute the `2`):

`I=2intu+12intdu+18int1/udu`

All of these are fairly common integrals:

`I=2(u^2/2)+12u+18lnabsu+C`

From `u=sqrtx-3` we see that:

`I=(sqrtx-3)^2+12(sqrtx-3)+18lnabs(sqrtx-3)+C`

Expanding the first two terms to simplify:

`I=(x-6sqrtx+9)+12sqrtx-36+18lnabs(sqrtx-3)+C`

The two constants can merge into `C`, the all-powerful constant of integration:

`I=x+6sqrtx+18lnabs(sqrtx-3)+C`