Thread: Definite Integrals for Integration by Parts

1. Definite Integrals for Integration by Parts

I'm having a bit of difficulty with this one. Can anyone help?
Many thanks.

Q. $\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2} logex dx Attempt: u = logex => du = \displaystyle \frac{1}{x} dx dv = \displaystyle \frac{1}{x^2} => dv = x-2 => v = \displaystyle \intx-2 => \displaystyle \frac{-x^-^2}{2} \displaystyle \intu dv = uv - \displaystyle \intv du => \displaystyle \int$$\displaystyle \frac{1}{x^2}$logex dx => logex$\displaystyle (\frac{-x^2}{2})$ + $\displaystyle \int$$\displaystyle \frac{x^-^2}{2}$$\displaystyle (\frac{1}{x})$ => $\displaystyle \frac{-x^-^2log_ex}{2}$ + $\displaystyle \int$$\displaystyle \frac{x^-^3}{2} => \displaystyle \frac{-x^-^2log_ex}{2} - \displaystyle \frac{x^-^2}{4} \displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$logex dx = $\displaystyle [\frac{-x^-^2log_ex}{2} - \frac{x^-^2}{4}]^2_1$ => $\displaystyle [\frac{-2^-^2log_e2}{2} - \frac{2^-^2}{4}] - [\frac{-1^-^2log_e1}{2} - \frac{1^-^2}{4}$ => $\displaystyle \frac{-log_e2}{8} - \frac{1}{16} + 0 + \frac{1}{4}$ => $\displaystyle \frac{-log_e2}{8} - \frac{5}{16}$

Ans.:
(From text book): $\displaystyle \frac{1}{2}$(1 - ln2)

2. Re: Definite Integrals for Integration by Parts

Check $\displaystyle \int x^{-2}dx$

3. Re: Definite Integrals for Integration by Parts

I've made the change I think you're referring to but I'm still off the mark...

u = logex => du = $\displaystyle \frac{1}{x}$ dx
dv = $\displaystyle \frac{1}{x^2}$ => => v = $\displaystyle \int$$\displaystyle \frac{1}{x^2} => ln(x2) \displaystyle \intu dv = uv - \displaystyle \intv du => \displaystyle \int$$\displaystyle \frac{1}{x^2}$logex dx => logex(ln(x2)) -$\displaystyle \int$ln(x2)$\displaystyle (\frac{1}{x})$ => ln(x2)logex - (x2ln(x2) - x2)$\displaystyle (\frac{1}{x})$
$\displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}logex dx = [ln(x2)logex - (x2ln(x2) - x2)\displaystyle (\frac{1}{x})]\displaystyle ^2_1 => [ln(4)loge2 - (4ln(4) - 4)\displaystyle (\frac{1}{2})] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4)\displaystyle (\frac{1}{2}) + 0 + 1 Ans.: (From text book): \displaystyle \frac{1}{2}(1 - ln2) 4. Re: Definite Integrals for Integration by Parts Originally Posted by GrigOrig99 I've made the change I think you're referring to but I'm still off the mark... u = logex => du = \displaystyle \frac{1}{x} dx dv = \displaystyle \frac{1}{x^2} => => v = \displaystyle \int$$\displaystyle \frac{1}{x^2}$ => ln(x2) no
$\displaystyle \int$u dv = uv - $\displaystyle \int$v du => $\displaystyle \int$$\displaystyle \frac{1}{x^2}logex dx => logex(ln(x2)) -\displaystyle \intln(x2)\displaystyle (\frac{1}{x}) => ln(x2)logex - (x2ln(x2) - x2)\displaystyle (\frac{1}{x}) \displaystyle \int^2_1$$\displaystyle \frac{1}{x^2}$logex dx = [ln(x2)logex - (x2ln(x2) - x2)$\displaystyle (\frac{1}{x})$]$\displaystyle ^2_1$ => [ln(4)loge2 - (4ln(4) - 4)$\displaystyle (\frac{1}{2})$] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4)$\displaystyle (\frac{1}{2})$ + 0 + 1

Ans.:
(From text book): $\displaystyle \frac{1}{2}$(1 - ln2)
$\displaystyle \int \frac{1}{x^2} \, dx = -\frac{1}{x} + C$

5. Re: Definite Integrals for Integration by Parts

Originally Posted by GrigOrig99
Ans.: [/B](From text book): $\displaystyle \frac{1}{2}$(1 - ln2)

And this.

6. Re: Definite Integrals for Integration by Parts

Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
Thanks again, guys.

7. Re: Definite Integrals for Integration by Parts

Originally Posted by GrigOrig99
Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
Thanks again, guys.
none ... in "big boy" math, log is understood base e