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

$$\int\ln(x+x^2)dx=x\ln(x+x^2)+\ln|x+1|-2x+C$$
$$A=\int\ln(x+x^2)dx=\int\ln\Big(x(x+1)\Big)dx$$ $$A=\int\Big(\ln x+\ln(x+1)\Big)dx$$ $$A=\int\ln xdx+\int\ln(x+1)dx$$ $$A=M+N$$ 1) First, we take a look at $$M=\int\ln xdx$$ Let $u=\ln x$ and $dv=dx$ We will have $du=\frac{1}{x}dx$ and $v=x$ Apply the formula $\int udv=uv-\int vdu$: $$M=x\ln x-\int x\times\frac{1}{x}dx$$ $$M=x\ln x-\int dx$$ $$M=x\ln x-x+C$$ 2) Now take a look at $$N=\int\ln (x+1)dx$$ Let $u=\ln (x+1)$ and $dv=dx$ We will have $du=\frac{1}{x+1}dx$ and $v=x$ Apply the formula $\int udv=uv-\int vdu$: $$N=x\ln (x+1)-\int \frac{x}{x+1}dx$$ $$N=x\ln (x+1)-\int \frac{x+1-1}{x+1}dx$$ $$N=x\ln(x+1)-\int\Big(1-\frac{1}{x+1}\Big)dx$$ $$N=x\ln(x+1)-\Big(\int1dx-\int\frac{1}{x+1}dx\Big)$$ $$N=x\ln(x+1)-\Big(x-\ln|x+1|\Big)+C$$ $$N=x\ln(x+1)+\ln|x+1|-x+C$$ Therefore, $A$ would be $$A=x\ln x+x\ln(x+1)+\ln|x+1|-2x+C$$ $$A=x\ln(x+x^2)+\ln|x+1|-2x+C$$