integration of (ln x)^2/ X^3 dx?

Question

question 28 james stewart chap 7.1

Answer 1

The repeated integration by parts works well,
but it is so long and tedious. Let 's find a simpler way.
Let's let u = ln x, x = e^u, dx = e^u du.
Plugging this into our integral reduces it to computing
∫ u² e^-(2u) du.
Since the 3rd derivative of u² is 0, we can use tic-tac-toe.
u² e^(-2u)
2u -e^(-2u)/2
2 e^(-2u)/4
0 -e^(-2u)/8
So the integral is equal to
-u² e^(-2u)/2 - ue^(-2u)/2 - e^-2u/4
Now back substitute to get the final answer:
∫ (ln x)² / x³ dx = -(ln x)² /2x² - (ln x)/2x² - 1/4x² + C.

Answer 2

2(Integral ln x * x^-3 dx)

ln x = u
x^-3 dx = dv
1/x dx = du
(-1/2)(1/x^2) = v

(-1/2)(ln x / x^2) - Integral (-1/2)(1/x^3) dx
(-1/2)(ln x / x^2) + (1/2)(-1/2)(1/x^2) + C
Multiply everything by 2.
(-ln x / x^2) - 1/(2x^2) + C

Answer 3

Try integration by parts
integ (ln x)^2 d(-1/2x^2)
dv=1/x^3 v=-(1/2x^2) u=(ln x)^2
uv-integ vdu
-(ln x)^2 /2x^2 + (1/2) integ 2ln(x)/x^3
=-(ln x)^2 / 2x^2 + integ ln(x)/x^3 -- (1)
consider integ ln(x)/x^3
integ ln(x) d(-1/2x^2)
apply uv-integ vdu again
-ln(x)/2x^2 + integ (1/2x^4)
=-ln(x) /2x^2 +(1/2)(-1/3x^3)
=-ln(x)/2x^2 -1/6x^3 -- (2)
(1)+(2)+c gives the answer.
I hope I did it OK.

Answer 4

[-2ln²(x) + 2ln(x) + 1] / 4x² + C