`(sqrt(1-x)+sqrt(1+x))^2`

`=1-x+1+x+2sqrt((1-x)(1+x))`

`=2(1+sqrt(1-x^2))`

So:

`"log" (sqrt(1-x)+sqrt(1+x))`

`= "log"((2(1+sqrt(1-x^2)))^(1/2))`

`= (1)/(2) ("log" 2 + "log" (1+sqrt(1-x^2) ))`

So:

`int ("log" (sqrt(1-x)+sqrt(1+x)) ) dx`

`=(1)/(2) int ("log" 2) dx + (1)/(2) int ("log" (1+sqrt(1-x^2))) dx `

The first integral is just constant:

`(1)/(2) int_0^1 ("log" 2) dx = (1)/(2) ("log" 2) (1-0) = ("log" 2)/(2)`

For the second integral, we start with:

`int_0^1 ("log" (1+sqrt(1-x^2)))dx `

We start with a trig substitution:

`x = "sin" theta`

`dx = "cos" theta d theta`

`1-x^2 = 1 - "sin"^2 theta = "cos"^2 theta`

`sqrt(1-x^2) = "cos" theta`

If `theta = 0` then `x=0`

If `theta = (pi)/(2)` then `x=1`

And the integral becomes:

`int_0^(pi/2) ("log" (1+ "cos" theta) "cos" theta) d theta`

Now we use integration by parts:

`u = "log" (1+ "cos" theta)`

`dv = "cos" theta d theta`

So:

`du = (-"sin" theta)/(1 + "cos" theta) d theta`

`v="sin" theta`

`int ("log" (1+ "cos" theta) "cos" theta) d theta`

`= "log" (1+ "cos" theta) "sin" theta - int (("sin" theta)(-"sin" theta)/(1 + "cos" theta)) d theta`

`= "log" (1+ "cos" theta) "sin" theta + int (("sin"^2 theta)/(1 + "cos" theta)) d theta`

Now:

`("sin"^2 theta)/(1+"cos" theta)`

`= (("sin"^2 theta)(1-"cos" theta))/((1+"cos" theta)(1-"cos" theta))`

`= (("sin"^2 theta)(1-"cos" theta))/(1-"cos"^2 theta)`

`= (("sin"^2 theta)(1-"cos" theta))/("sin"^2 theta)`

`=1-"cos" theta`

Thus our integral becomes:

`= "log" (1+ "cos" theta) "sin" theta + int (1-"cos" theta) d theta`

`= "log" (1+ "cos" theta) "sin" theta + theta - "sin" theta |_0^(pi/2)`

`int_0^1 ("log" (sqrt(1-x)+sqrt(1+x)) ) dx`

` = ("log" 2)/(2) + (1)/(2)( "log" (1+ "cos" theta) "sin" theta + theta - "sin" theta |_0^(pi/2) )`

` = ("log" 2)/(2) + (1)/(2)( "log" (1+ 0) *1 + (pi)/(2) - 1 ) - (1)/(2)( "log" (1+ 1) *0 + 0 - 0 )`

` = ("log" 2)/(2) + (1)/(2)( (pi)/(2) - 1 ) `

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