## Calculus: Early Transcendentals 8th Edition

$$\int\frac{x}{x^2+4}dx=\frac{\ln(x^2+4)}{2}+C$$
$$A=\int\frac{x}{x^2+4}dx$$ Let $u=x^2+4$. We would have $du=2xdx$. Then it can be deduced that $xdx=\frac{1}{2}du$ Substitute into $A$, we have $$A=\frac{1}{2}\int\frac{1}{u}du$$ $$A=\frac{1}{2}\ln|u|+C$$ $$A=\frac{\ln|x^2+4|}{2}+C$$ But here we also notice that as for all $x\in R$, $x^2\geq0$. Therefore, $x^2+4\gt0$ for all $x\in R$ as well. That means $|x^2+4|=x^2+4$. Therefore, $$A=\frac{\ln(x^2+4)}{2}+C$$