Integration by parts involving exponentials and logarithms

xllx
Messages
33
Reaction score
0

Homework Statement



Using integration by parts, integrate:
(1/x^2)(lnx) dx with the limits e and 1


Homework Equations



[uv]to the limits a b - the integral of (v)(du/dx) dx

(sorry, don't know how to write out equations properly on a computer)

The Attempt at a Solution



I've got u=x^-2 so du/dx= -2x^-3
dv/dx=lnx so v= x(lnx-1)

So putting this into the equation above:
[x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1))

Is this right so far?
If so how do I integrate the last part? Do I do it sepeartley or by parts again?

Many Thanks, any help at all would be greatly appreciated.
 
Physics news on Phys.org
Hi xllx! :smile:

(have an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
xllx said:
Using integration by parts, integrate:
(1/x^2)(lnx) dx with the limits e and 1

I've got u=x^-2 so du/dx= -2x^-3
dv/dx=lnx so v= x(lnx-1)

So putting this into the equation above:
[x^-2.x(lnx-1)] - the integral of (-2x^-3.x(lnx-1))

eugh!

go the other way …

integrate x-2 ! :smile:
 
Thankyou. Redid it and came out with a reasonal answer.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top