Question

Question #fcb8e

Answer 1

`(6arctan(4/sqrt3)-pi)/(12sqrt3)approx0.18434`

Explanation:

`I=int_0^(ln2)e^(2x)/(e^(4x)+3)dx`

Let `u=e^(2x)`. This implies that `du=2e^(2x)dx`, which we are off of by a factor of `2`. We will also change the bounds by plugging `x=0,ln2` into `u=e^(2x)`, giving bounds of `u=e^0=1` and `u=e^(2ln2)=e^ln4=4`.

`I=1/2int_0^(ln2)(2e^(2x))/((e^(2x))^2+3)dx=1/2int_1^4(du)/(u^2+3)`

We can manipulate this into the arctangent integral:

`I=1/6int_1^4(du)/(u^2/3+1)=1/6int_1^4(du)/((u/sqrt3)^2+1)`

Let `v=u/sqrt3` so `dv=1/sqrt3du`. Don't forget to change the bounds:

`I=sqrt3/6int_1^4(1/sqrt3du)/((u/sqrt3)^2+1)=1/(2sqrt3)int_(1/sqrt3)^(4/sqrt3)(dv)/(v^2+1)`

This is the arctangent integral:

`I=1/(2sqrt3)arctan(v)|_(1/sqrt3)^(4/sqrt3)=1/(2sqrt3)(arctan(4/sqrt3)-arctan(1/sqrt3))`

`I=arctan(4/sqrt3)/(2sqrt3)-1/(2sqrt3)(pi/6)`

`I=(6arctan(4/sqrt3)-pi)/(12sqrt3)approx0.18434`