How do I integrate 1/((x^(1/2))+1)? it appears to be integrate by parts but I can't seem to pick up the catch.

Answer 1

It's not "integrate by parts."

It's tailored for making the substitution x = y^2, ### and then using "partial fractions" :

Then dx / [x^(1/2) + 1] = 2 y dy / [y + 1] = [2 y + 2 - 2] dy / [y + 1]

= 2 [1 - 1 / (y + 1)] dy.

The integral is then 2 [y - ln (y + 1)] + C

= 2 [x^(1/2) - ln (x^(1/2) + 1)] + C.

Live long and prosper.

### Often, when an x^(1/2) makes an appearance in an integral wrt x, it cries out for this substitution x = y^2. The x^(1/2) terms then become linear in ' y ,' which is generally much easier to deal with.

Answer 2

You don't use parts here. Instead let x = u², dx = 2u du.
Then you have to integrate
2u/(u+1) . Write this as 2( 1 - 1/(u+1) )
and the integral works out to be
2u -2 log(u+1) + C = 2√x - 2 log(√x + 1) + C.

Answer 3

Substitute u^2 for x, 2u du for dx, to get

integral of 2u du / (u + 1)

Then 2u / (u + 1) = (2u + 2 - 2) / (u + 1) = (2 - (2 / (u + 1)))

which integrates to 2(u - ln(u + 1)) + C or 2(sqrt(x) - ln(sqrt(x) + 1)) + C

Dan

Answer 4

1/((sqrt x)+1) =
2(1+sqrt(x) – 2 ln(1+ sqrt (x)) +C
= 2(sqrt (X) – ln (1 + sqrt (x)) + C