Question #fcb8e

1 Answer
Feb 6, 2017

#(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#