How do you integrate #x/(1+e^(2x^2))#?
1 Answer
Mar 22, 2015
We're going to need some substitutions, I'll make one at the time to be as clear as possible:
- First of all, you substitute
#u=x^2# , to get#du=2xdx# and transform the integral into
#1/2 \int{1}/{e^{2u}+1}du# - Then, we'll go with
#s=2u# . This means#ds=2du# , so we only have to add another factor#1/2# , and get
#1/4 \int{1}/{e^{s}+1}ds# - Now, let's get rid of the exponential, defining
#p=e^s# . This means#dp=e^sds# . Multiplying and dividing by#e^s# , we get
#1/4 \int{1}/{p(p+1)}dp# - Writing
#1/{p(p+1)}# as#1/p - 1/(p+1)# , we can split the integrals, and get
#-1/4\int 1/{p+1} dp + 1/4\int 1/p dp# . - These are both known integral, giving
#log(p+1)# and#\log(p)# . Now we need to make all the substitutions backwards: from
#-1/4\log(p+1)+ 1/4\log(p)# #p=e^s# , so we have
#-1/4 \log(e^s+1) + 1/4\log(e^s)# #s=2u# , so we have
#-1/4 \log(e^{2u}+1) + 1/4\log(e^{2u})# - Finally,
#u=x^2# , so we have
#-1/4 \log(e^{2x^2}+1) + 1/4\log(e^{2x^2})#
The only further step we can do is writing the answer as
since we have
