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

$\ln |1+\sqrt x|+c$
Solve $\int\dfrac{dx}{2 \sqrt x+2x}$ or, $\int\dfrac{dx}{2 \sqrt x+2x}=\int\dfrac{dx}{2 \sqrt x(1+\sqrt x)}$ Let $p=1+\sqrt x$ and $dp=\dfrac{1}{2 \sqrt x }dx$ This implies $\int\dfrac{dx}{2 \sqrt x(1+\sqrt x)}=\int\dfrac{dp}{p}=\ln|p|+c$ Since, $p=1+\sqrt x$ Thus, $\int\dfrac{dx}{2 \sqrt x+2x}=\ln |1+\sqrt x|+c$