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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

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

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

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

$C \in \mathbb{R}$

Answer 4

$$ \begin{aligned} \int \frac{\sqrt{x+25}}{x} d x &=\int \frac{x+25}{x \sqrt{x+25}} d x \\ &=2 \int \frac{x+25}{x} d(\sqrt{x+25}) \\ &=2 \int 1 d(\sqrt{x+25})+50 \int \frac{d(\sqrt{x+25})}{(\sqrt{x+25})^{2}-5^2}\\ &=2 \sqrt{x+25}+10 \ln \left|\frac{\sqrt{x+25}-5}{\sqrt{x+25}+5}\right|+C \end{aligned} $$