Verified answer

Use Integration by Parts

Let u = x and dv = e^x dx, then du = dx and v = e^x

u*v - ⌠ v du

xe^x - ⌠ e^x dx

xe^x - e^x + C <==== Solution

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

Check solution:

Take the derivative of x*e^x - e^x + C,

e^x + x*e^x - e^x

= e^x - e^x + x*e^x

= 0 + x*e^x

= x*e^x <==== This matches the integrand.

∫ [1/(x(x^4-16)^(1/2))] dx =

∫ [x / (x^2 (x^4-16)^(1/2) )] dx =

(1/2)∫ [1 / (x^2 (x^4-16)^(1/2) )] d(x^2 ) or use x^2 = u you get

(1/2)∫ 1 / (u (u^2-16)^(1/2) ) du =

(1/2)∫ 1 / (u^2 (1-(4/u)^2)^(1/2) ) du =

- (1/8)∫ 1 /√(1-(4/u)^2) d(4/u) =

(-1/8)Arcsin(4/u) + C

(-1/8)Arcsin(4/x^2) + C

Arcsin or sin^(-1)