How do you find the indefinite integral of #int (19dx)/ [ x^2 * sqrt(x^2+62) ]#?

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Answer 1 · 2018-03-10T13:48:51

The answer is #=-19/62coth(arcsinh(x/sqrt62))+C#

We need

#intcsch^2xdx=-cothx#

The integral is

#I=int(19dx)/(x^2sqrt(x^2+62))=19int(dx)/(x^2sqrt(x^2+62))#

Perform the substitution

#x=sqrt62sinhu#, #=>#, #dx=sqrt62coshu#

#x^2+62=62sinh^2u+62=62cosh^2u#

Therefore,

#I=19int(sqrt62coshudu)/(62sinh^2u*sqrt62coshu)#

#=19/62int(du)/(sinh^2u)#

#=19/62int(csch^2udu)#

#=-19/62cothu#

#=-19/62coth(arcsinh(x/sqrt62))+C#