How I resolve this integral?

Question

#int(x+sqrt(x+1))/(2x+1)dx#

I try to substitute #sqrt(x+1)=t#

And the result is this:
#t+t^2/2-sqrt(2)/4ln|sqrt(2)-2t|+sqrt(2)/4ln|sqrt(2)+2t|-1/2ln|2t^2-1| + c# where #t=sqrt(x+1)#

It's right?

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Answer 1 · 2018-06-22T13:41:20

#1/4*t^2+1/2t-1/8*ln(2t^2-1)-1/4*sqrt(2)arc tanh(t*sqrt(2))+C#
where #t=sqrt(x+1)#

With #t=sqrt(x+1)# we get #x=t^2-1# and

#dx=2tdt#
so our integral will be
#2*int (t^3+t^2-1)/(2t^2-1)dt#

converting into partial factions

#int (t+1+(1-t)/(2t^2-1))dt#
this gives

#1/4t^2+1/2t-1/8ln(2t^2-1)-1/4sqrt(2)*arc tanh(tsqrt(2))+C#