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Answer 1 · 2017-02-09T23:09:24

#int (x^4dx)/sqrt(x^10-2) = 1/5 ln (x^5/sqrt2 + sqrt(x^10/2-1)) +C#

Evaluate:

#int (x^4dx)/sqrt(x^10-2)#

Substitute:

#t=x^5#
#dt =5x^4#

#int (x^4dx)/sqrt(x^10-2) = 1/5int (dt)/sqrt(t^2-2)#

Substitute:

#t =sqrt2 cosh u#

#dt = sqrt2 sinh u#

#1/5int (dt)/sqrt(t^2-2) =sqrt2/5 int (sinh u du)/sqrt(2cosh^2-2)#

Use:

#cosh^2u -1 = sinh^2u#

#1/5 int sinh u/sqrt(sinh^2u)du = 1/5 int du = 1/5 u +C#

Reversing the substitutions:

#int (x^4dx)/sqrt(x^10-2) = 1/5arccosh(x^5/sqrt2) + C#

or using the logarithmic form of #arccosh#:

#int (x^4dx)/sqrt(x^10-2) = 1/5 ln (x^5/sqrt2 + sqrt(x^10/2-1)) +C#