# Need help getting started with a partial fraction $\int\frac{\sqrt{x+25}}{x}$

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.

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

• How did you know to make that substitution? – hax0r_n_code Sep 30 '13 at 0:20
• @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. – awwalker Sep 30 '13 at 0:23
• 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. – André Nicolas Sep 30 '13 at 0:26
• @inquisitor well, you have to be careful, I missed at first that Andre set $u^2=x+25$, so $du$ isn't $dx$ – Jean-Sébastien Sep 30 '13 at 0:41
• 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. – André Nicolas Sep 30 '13 at 0:47

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

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

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

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

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

$C \in \mathbb{R}$

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