How do I evaluate #intsqrt(36+9x^2)dx#?

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Answer 1 · 2015-02-15T19:16:47

The answer is: #3/2*xsqrt(4+x^2)+6arcsinh(x/2)+c#

First of all:

#intsqrt(36+9x^2)dx=3intsqrt(4+x^2)dx# and now we have to substitute:

#x=2sinhtrArrdx=2coshtdt#.

So:

#3intsqrt(4+x^2)dx=3intsqrt(4+4sinh^2t)*2coshtdt=#

#=3int2sqrt(1+sinh^2t)*2coshtdt=12intcosh^2tdt=#

#=12int(cosh2t+1)/2dt=6(sinh2t)/2+6t+c=#

#=3*2sinhtcosht+6t+c=#

Now: #sinht=x/2#, #t=arcsinh(x/2)#

since #cosht=sqrt(1+sinh^2t#, than #cosht=sqrt(1+x^2/4)=#

#=sqrt((4+x^2)/4)=1/2sqrt(4+x^2)#

So:

#I=6*x/2*1/2sqrt(4+x^2)+6arcsinh(x/2)+c=#

#=3/2*xsqrt(4+x^2)+6arcsinh(x/2)+c#