What is integral of [x^2+1/{(x^2+4)^0.5}]?

1 Answer
Jul 6, 2018

#-ln|x+sqrt(x^2+4)|+x/2sqrt(x^2+4)+C#

Explanation:

I assume w.r.t. x...
#I=int(x^2+1)/sqrt(x^2+4)dx#

Substitute #x=2tanw#. Then #dx/(dw)=2sec^2w# and
#I=int(1+4tan^2w)/sqrt(4+4tan^2w)*2sec^2wdw#
#I=int(1+4tan^2w)/(2secw)*2sec^2wdw#
#I=int(1+4tan^2w)secwdw#
#I=intsecw+4secwtan^2wdw#
#I=intsecw+4secw(sec^2w-1)dw#
#I=int-3secw+4sec^3wdw#

The integral of #secxdx# is #ln|secx+tanx|#; see
https://socratic.org/questions/what-is-the-integral-of-sec-x

The integral of #sec^3xdx# is #1/2secxtanx+1/2ln|secx+tanx|#; see
https://socratic.org/questions/what-is-the-integral-of-sec-3-x

So
#I=-3ln|secw+tanw|+2secwtanw+2ln|secw+tanw|+C#
#I=-ln|secw+tanw|+2secwtanw+C#

Substitute back for our original variable #x#, using #tanw=x/2#
and #secw=sqrt(1+x^2/4)#:

#I=-ln|sqrt(1+x^2/4)+x/2|+xsqrt(1+x^2/4)+C#
#I=-ln(1/2|x+sqrt(x^2+4)|)+x/2sqrt(x^2+4)+C#
#I=-ln|x+sqrt(x^2+4)|+x/2sqrt(x^2+4)+C#
where we have absorbed the factor inside the log into the integration constant in order to make our expression similar to the question expression.