Question #15df8

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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 $arccosh$ :

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