Evaluate the definite integral. ?

Question

Integral from e^4 to e^16 of

(dx)/(xsqrt(ln(x))

Please explain. Thanks

Answer 1

we will do a "u" substitution to solve this integral

u = lnx

du = (1/x)dx

dx = x du

Substitute in and we get:

Integral from e^4 to e^16 of

(x du)/(xsqrt(u))

the x's cancel out and i will rewrite 1/sqrt(u) as u^-1/2

Integral from e^4 to e^16 of

u^(-1/2)du

integrate by using reverse power rule

u^1/2/(1/2) evaluated from e^4 to e^16

2sqrt(lnx) evaluated from e^4 to e^16

2sqrt(ln e^16) - 2sqrt(ln e^4)

2sqrt(16) - 2sqrt(4)

2(4) - 2(2)

8 - 4 = 4

voila

Answer 2

∫dx/(x√ln(x))

let u = ln(x) ; du/dx = 1/x

∫dx/(x√ln(x)) = ∫du/√u = 2√u + c = 2√ln(x) + c

2√ln(e^16) - 2√ln(e^4) =  8 - 4 = 4

Answer: 4

  

Answer 3

Substitute x = e^y

Then we have integral from 4 to 16 of e^y dy / (e^y sqrt(y))

= Int dy/sqrt(y) = (2/3)y^(3/2)

Top limit: (2/3) 16^(3/2) = (2/3) 4^3 = (2/3) 64

Lower limit: (2/3) 4^(3/2) = (2/3) 2^3 = (2/3) 8

So the integral is (2/3)(64 - 8) = (2/3) * 56