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

$$\int^{4}_{2}x^{2x}(1+\ln x)dx=32760$$
$$A=\int^{4}_{2}x^{2x}(1+\ln x)dx$$ We would prove that $$(x^{2x})'=2(1+\ln x)x^{2x}$$ We have $$D= x^{2x}$$ Take the natural logarithm of both sides. $$\ln D=\ln(x^{2x})=2x\ln x$$ Therefore, $$\frac{d(\ln D)}{dx}=2(x'\ln x+x(\ln x)')$$ $$\frac{1}{D}\times\frac{dD}{dx}=2(\ln x+1)$$ $$\frac{dD}{dx}=2(\ln x+1)D=2(\ln x+1)x^{2x}$$ $$(x^{2x})'=2(1+\ln x)x^{2x}$$ That means, $$\int(1+\ln x)x^{2x}dx=\frac{1}{2}(x^{2x})+C$$ So, $$A=\frac{1}{2}(x^{2x})\Big]^4_2$$ $$A=\frac{1}{2}(4^8-2^4)=32760$$