Question

How do you evaluate the integral `int (1-sqrtx)/(1+sqrtx)`?

Answer 1

`-x + 4(1+ sqrt(x)) - 4ln|1 + sqrt(x)| + C`

Explanation:

Use partial fractions to get rid of the radical in the numerator.

`A/1 + B/(1 + sqrt(x)) = (1 - sqrt(x))/(1 + sqrt(x))`

`A(1 + sqrt(x)) + B = 1 - sqrt(x)`

`A + Asqrt(x) + B = 1 - sqrt(x)`

This means that `{(A = -1), (A + B = 1):}`

Solving, we get that `A = -1` and `B = 2`. The integral becomes

`int -1 + 2/(1 + sqrt(x))dx`

`int -1dx + int 2/(1 + sqrt(x))dx`

`int -1dx + 2int 1/(1 + sqrt(x))dx`

Let `u = 1 + sqrt(x)`. Then `du = 1/(2sqrt(x))dx` and `dx = 2sqrt(x)du`. We can also conclude that `2sqrt(x) = 2(u - 1)`, because `sqrt(x) = u -1` and we have two of those.

`int -1dx + 2int 1/u * 2(u - 1)du`

`int -1dx + 4int (u - 1)/u du`

`int -1dx + 4int u/udu - 4int 1/udu`

`int -1dx + 4int 1du- 4int 1/u du`

`-x + 4u - 4ln|u| + C`

`-x + 4(1+ sqrt(x)) - 4ln|1 + sqrt(x)| + C`

Hopefully this helps!