3
$\begingroup$

I'm having trouble determining how to get started with this problem. I tried rationalizing the fraction, but I didn't think that was correct after I got started. Here is the problem:

$$\int\frac{\sqrt{x+25}}{x}$$

Edit, not looking for someone to solve this for me, just initial support with getting it going.

$\endgroup$
1
$\begingroup$

A small start: Let $u^2=x+25$. We end up integrating a rational function. After a while, partial fractions will be involved.

$\endgroup$
  • $\begingroup$ How did you know to make that substitution? $\endgroup$ – hax0r_n_code Sep 30 '13 at 0:20
  • 1
    $\begingroup$ @inquisitor Because the square root is the worst part of the integrand. Any $u$-substitution that simplifies the problem is a good place to start. $\endgroup$ – awwalker Sep 30 '13 at 0:23
  • $\begingroup$ An old joke, one calls it a $U$-substitution because we substitute for the Ugly part. Don't like square roots, let's get rid of it. $\endgroup$ – André Nicolas Sep 30 '13 at 0:26
  • 1
    $\begingroup$ @inquisitor well, you have to be careful, I missed at first that Andre set $u^2=x+25$, so $du$ isn't $dx$ $\endgroup$ – Jean-Sébastien Sep 30 '13 at 0:41
  • 1
    $\begingroup$ We have $dx=2u\,du$, so we are integrating $\frac{2u^2}{u^2-25}$. Divide. We get $2+\frac{50}{u^2-25}$. Now partial fractions. $\endgroup$ – André Nicolas Sep 30 '13 at 0:47
1
$\begingroup$

Do a $u$ substitution of $u = \sqrt{x + 25}$ and then try long division on the result.

$\endgroup$
1
$\begingroup$

By The substitution $t^2 = x + 25 \implies 2tdt = dx $, then we obtain:

$$ \int \frac{\sqrt{x + 25}}{x} = - 2 \int \frac{t^2}{t^2 - 25} = -2 \int \frac{t^2 - 25 + 25}{t^2 - 25} $$

$$ = -2 \int dt -2\int\frac{25}{t^2 - 5^2} = -2t -50 \int \frac{1}{t^2 - 5^2} = -2t -50 \frac{1}{-25} Arctan(t) + C $$

$$ = -2t + 2 Arctan (t) + C = -2 \sqrt{x + 25} + 2 Arctan( \sqrt{x + 25 } ) + C$$

$C \in \mathbb{R}$

$\endgroup$
  • $\begingroup$ Where you see arctan, I see partial fraction $\endgroup$ – Jean-Sébastien Sep 30 '13 at 0:31
  • $\begingroup$ great ! :) :) :) $\endgroup$ – ILoveMath Sep 30 '13 at 0:32
  • $\begingroup$ I said that so you'd change it, you dont get arctan $\endgroup$ – Jean-Sébastien Sep 30 '13 at 2:02

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.