How to integrate #intu/(u^2+3/4) du# ?

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Answer 1 · 2018-07-02T18:00:01

#intu/(u^2+3/4)du=1/2int(2u)/(u^2+3/4)du=1/2int(d(u^2+3/4))/(u^2+3/4)du=1/2ln|u^2+3/4|+c#

Here,

#I=intu/(u^2+3/4)du#

Subst. #u^2+3/4=x=>2udu=dx=>udu=1/2dx#

So,

#I=int1/x*1/2dx#

#=>I=1/2int1/xdx#

#=>I=1/2ln|x|+c#

Subst. back #x=u^2+3/4#

#I=1/2ln|u^2+3/4|+c#