1) int{(2e^x - 2e^-x)/[(e^x + e^-x)^2]}dx

We will do this by substitution.

u = e^x + e^-x

=> du = e^x - e^-x dx

so our integral becomes

2*int{1/(u^2)}du

= -2(u)^-1 + C

= -2/(e^x + e^-x) + C

2) int{(x^2)/(x^2 +4)}dx

= int{(x^2 + 4 - 4)/(x^2 + 4)}dx

= int{1 + 4/(x^2 + 4)}dx

= int{1 + 4/[4((x/2)^2 + 1)]}dx

= x - 2arctan(x/2) + C ............I skipped a step, to get here i used u = x/2 substitution.

4) int{(cosx)/(1+(sinx)^2)}dx

let u = sinx

=> du = cosx dx

so our integral becomes

int{1/(1 + u^2)}du

= arctanu + C

= arctan(sin(x)) + C

evaluating between pi/4 and pi/2 we get

arctan(sin(pi/2)) - arctan(sin(pi/4))

= arctan1 - arctan(sqrt(2)/2)

i don't rmember the exact value for these.

3) Still working on this, I'm pretty sure we have to use trig substituion, which i'm a little rusty on, but there may be another way. I'm in a rush, so i'll type up to where i got and then leave the rest to you.

int{2x/sqrt(4x - x^2)}dx

= int{2x/sqrt[-(x^2 - 4x + (-2)^2 - (-2)^2)]}dx...............i'm completing the square here

= int{2x/sqrt[4 - (x - 2)^2]}dx

= int{2x/sqrt[2^2 - (x - 2)^2]}dx

now we apply the trigonometric substitutions.

we let x - 2 = 2sin(u)

so dx = 2cos(u) du....and i'm late for class, catch you later