## University Calculus: Early Transcendentals (3rd Edition)

$3 \ln |x|- \dfrac{x^{2}}{2}+c$
Calculate the anti-derivative. Since, we know $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+c$ and $\int \dfrac{1}{x} dx=\ln |x|+c$ where $c$ is a constant of proportionality. Then $\int (\dfrac{3}{x}-x) dx=3 \int \dfrac{1}{x} dx-\int x dx dx$ or, $=3 \ln |x|- \dfrac{x^{1+1}}{1+1}+c$ Thus, $=3 \ln |x|- \dfrac{x^{2}}{2}+c$