1
$\begingroup$

Show that $$\int_x^1\frac{dt}{1+t^2}=\int_1^{1/x}\frac{dt}{1+t^2},\;x>0.$$

So, I'm learning about integration techniques, and I get this exercise. We've been practicing $u$-substitution, and I think that might be involved here somehow, but I don't understand what I need to do to proceed. Thank you for your help!

$\endgroup$
3
  • $\begingroup$ Your bounds on the second integral are in the wrong order: the integral on the left is positive and that on the right is negative. $\endgroup$ Feb 5, 2014 at 1:37
  • $\begingroup$ Why was this question down-voted? $\endgroup$ Feb 5, 2014 at 1:40
  • $\begingroup$ @MichaelHardy I was confused about that. I've transcribed the problem (Apostol's Calculus, 5.8.23) correctly, so the book must be erroneous. Thank you for your help! $\endgroup$
    – Jackson
    Feb 5, 2014 at 1:49

2 Answers 2

2
$\begingroup$

$$ \begin{align} u & = \frac 1 t \\[8pt] du & = \frac{-dt}{t^2} \\[8pt] -t^2\,du & = dt \\[8pt] \frac{-du}{u^2} & = dt \end{align} $$ When $t=x$ the $u=1/x$.

So $$ \int_1^x \frac{dt}{1+t^2} = \int_1^{1/x} \frac{-du/u^2}{1+(1/u)^2} = \int_1^{1/x} \frac{-du}{u^2+1}. $$

$\endgroup$
2
$\begingroup$

the substitution $t \leftarrow 1/t$ yields $$\int_1^x {dt\over 1 + t^2 } = \int_{1/x}^1 {dt/t^2\over 1 + (1/t)^2} = \int_{1/x}^1 {dt\over 1 + t^2}. $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.