# Math Help - 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. $\int^2_1$ $\frac{1}{x^2}$ logex dx

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

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

2. ## Re: Definite Integrals for Integration by Parts

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

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 = $\frac{1}{x}$ dx
dv = $\frac{1}{x^2}$ => => v = $\int$ $\frac{1}{x^2}$ => ln(x2)
$\int$u dv = uv - $\int$v du => $\int$ $\frac{1}{x^2}$logex dx => logex(ln(x2)) - $\int$ln(x2) $(\frac{1}{x})$ => ln(x2)logex - (x2ln(x2) - x2) $(\frac{1}{x})$
$\int^2_1$ $\frac{1}{x^2}$logex dx = [ln(x2)logex - (x2ln(x2) - x2) $(\frac{1}{x})$] $^2_1$ => [ln(4)loge2 - (4ln(4) - 4) $(\frac{1}{2})$] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4) $(\frac{1}{2})$ + 0 + 1

Ans.:
(From text book): $\frac{1}{2}$(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 = $\frac{1}{x}$ dx
dv = $\frac{1}{x^2}$ => => v = $\int$ $\frac{1}{x^2}$ => ln(x2) no
$\int$u dv = uv - $\int$v du => $\int$ $\frac{1}{x^2}$logex dx => logex(ln(x2)) - $\int$ln(x2) $(\frac{1}{x})$ => ln(x2)logex - (x2ln(x2) - x2) $(\frac{1}{x})$
$\int^2_1$ $\frac{1}{x^2}$logex dx = [ln(x2)logex - (x2ln(x2) - x2) $(\frac{1}{x})$] $^2_1$ => [ln(4)loge2 - (4ln(4) - 4) $(\frac{1}{2})$] - ln(1)loge1 - (1ln(1) - 1)(1)] => ln(4)loge2 - (4ln(4) - 4) $(\frac{1}{2})$ + 0 + 1

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

5. ## Re: Definite Integrals for Integration by Parts

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