Question

How do you integrate `int_0^1 1/(1+(x)^(1/2)) dx` ?

Answer 1

Write `x^(1/2)` as `sqrtx`:

`int_0^1 1/(1+sqrtx) dx`

To do the indefinite integral, I used a tip from WolframAlpha :

Let `u = sqrtx` then `du = 1/(2sqrtx)dx`

We need to solve for `dx`:

`dx = 2sqrtxdu`

But `sqrtx = u`:

`dx = 2udu`

Perform the substitutions:

`int 1/(1+(x)^(1/2)) dx = 2 int u/(u+1) du`

Add 0 to the numerator in the form if `+1 - 1`

`int 1/(1+(x)^(1/2)) dx = 2 int (u+1-1)/(u+1) du`

Separate into two fractions:

`int 1/(1+(x)^(1/2)) dx = 2 int (u+1)/(u+1)-1/(u+1) du`

The first fraction becomes 1:

`int 1/(1+(x)^(1/2)) dx = 2 int 1-1/(u+1) du`

Both terms are trival to integrate:

`int 1/(1+(x)^(1/2)) dx = 2 (u-ln(u+1))+C`

Reverse the substitution:

`int 1/(1+(x)^(1/2)) dx = 2 (sqrtx-ln(sqrtx+1))+C`

To integrate from 0 to 1 we evaluate the right side at `x = 1` and subtract the right side evaluated at `x = 0`:

`int_0^1 1/(1+(x)^(1/2)) dx = {2(sqrt1-ln(sqrt1+1))+C}-{2(sqrt0-ln(sqrt0+1))+C}`

The constants of integration sum to 0 and `sqrt0 - ln(1) = 0`, therefore, the answer is:

`int_0^1 1/(1+(x)^(1/2)) dx = 2-2ln(2)`

Answer 2

`f(x) = 1/(1+sqrtx)`

let `sqrtx=t`
`x=t^2`
differentiating both the sides,
`dx = 2t.dt`

`intf(x) = int1/(1+sqrtx) dx`
`= int1/(1+t) 2t. dt = int(2t)/(1+t) dt`
applying simple polynomial division,
`int (2-(2)/(1+t)) dt`
`int 2.dt-int(2)/(1+t) dt`
`2intdt-2int1/(1+t) dt`
`2t-2ln(1+t) +c`

replacing, `t=sqrtx`
`2sqrtx-2ln(1+sqrtx) +c`

now, applying the limits,
`2sqrt1-2ln(1+sqrt1) - (2sqrt0-2ln(1+sqrt0))`
`2-2ln(1+1) - (-2ln(1))` as `color(purple)(sqrt1=1; sqrt0=0;ln1=0`
`2-2ln2`