# Find the general anti-derivative of {9 + square root x} / x.

## Question:

Find the general anti-derivative of {eq}\displaystyle \frac{9+\sqrt{x}}{x} {/eq}.

The general antiderivative of {eq}\displaystyle \frac{9+\sqrt{x}}{x} {/eq} corresponds to the solution of {eq}\displaystyle \int \displaystyle \frac{9+\sqrt{x}}{x} \ \mathrm{d}x {/eq}:

{eq}\begin{align*} \displaystyle \int \displaystyle \frac{9+\sqrt{x}}{x} \ \mathrm{d}x & =\displaystyle \int \displaystyle \left( \frac{9}{x}+ x^{-\frac{1}{2}}\right) \ \mathrm{d}x\\ & =\displaystyle \int \frac{9}{x} \ \mathrm{d}x+ \int x^{-\frac{1}{2}} \ \mathrm{d}x \ \ \ \left[\mathrm{ Utilize \ the \ sum \ rule \ for \ integrals }\right]\\ & = 9 \ln |x| + \int x^{-\frac{1}{2}} \ \mathrm{d}x \ \ \ \left[\mathrm{ Apply \ the \ integration \ formula: \ }\int \frac{\mathrm{d}u}{u}=\ln |u| +C\right]\\ & = 9 \ln |x| + \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C \ \ \ \left[\mathrm{ Use \ the \ power \ rule \ for \ integrals: \ }\displaystyle \int x^n \ \mathrm{d}x = \frac{x^{n+1}}{n+1}+C, C \ \mathrm{is \ the \ constant \ of \ integration} \right]\\ \implies \displaystyle \int \displaystyle \frac{9+\sqrt{x}}{x} \ \mathrm{d}x& = 9 \ln |x| + 2\sqrt{x}+C\\ \end{align*} {/eq}

Hence, {eq}9 \ln |x| + 2\sqrt{x}+C {/eq} is the general anti-derivative of {eq}\displaystyle \frac{9+\sqrt{x}}{x} {/eq}.