Please show me how to integrate intergal! ?

Question

e^36

_____x_____ dx

x[ (lnx)^1/2]

e^9

Answer 1

∫ dx/[x (lnx)^(1/2)] =

rewrite the integrand as:

∫ [(lnx)^(-1/2)] (1/x) dx =

and note that it includes both the function lnx and its derivative (1/x);

therefore let:

lnx = u

differentiate both sides:

d(lnx) = du →

(1/x) dx = du

thus, substitute, yielding:

∫ [(lnx)^(-1/2)] (1/x) dx = ∫ [u^(-1/2)] du =

{u^[(-1/2)+1]} /[(-1/2)+1] + C =

[u^(1/2)] /(1/2) + C =

2u^(1/2) + C

thus, being u = lnx, the antiderivative is:

2 (lnx)^(1/2) + C

finally, evaluating the definite integral from e^9 to e^36, you get:

2 [ln(e^36)]^(1/2) - 2 [ln(e^9)]^(1/2) =

2 [36^(1/2)] - 2 [9^(1/2)] =

2(6) - 2(3) =

12 - 6 = 6

I hope it helps..

Bye!