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

$$\int\frac{(x+1)^2\tan^{-1}(3x)+9x^3+x}{(9x^2+1)(x+1)^2}dx=\frac{1}{6}(\tan^{-1}(3x))^2+\ln|x+1|+\frac{1}{x+1}+C$$
$$A=\int\frac{(x+1)^2\tan^{-1}(3x)+9x^3+x}{(9x^2+1)(x+1)^2}dx$$ $$A=\int\frac{(x+1)^2\tan^{-1}(3x)}{(9x^2+1)(x+1)^2}dx+\int\frac{x(9x^2+1)}{(9x^2+1)(x+1)^2}dx$$ $$A=\int\frac{\tan^{-1}(3x)}{9x^2+1}dx+\int\frac{x}{(x+1)^2}dx$$ $$A=M+N$$ 1) Consider $M$: $$M=\int\frac{\tan^{-1}(3x)}{9x^2+1}dx$$ Set $u=\tan^{-1}(3x)$, which means $$du=\frac{(3x)'}{(3x)^2+1^2}dx=\frac{3}{9x^2+1}dx$$ $$\frac{1}{9x^2+1}dx=\frac{1}{3}du$$ Therefore, $$M=\frac{1}{3}\int udu$$ $$M=\frac{1}{6}u^2+C$$ $$M=\frac{1}{6}(\tan^{-1}(3x))^2+C$$ 2) Consider $N$: $$N=\int\frac{x}{(x+1)^2}dx=\int\Big(\frac{A}{x+1}+\frac{B}{(x+1)^2}\Big)dx$$ We now clear the fractions. $$A(x+1)+B=x$$ $$Ax+A+B=x$$ So, $A=1$ while $A+B=0$, meaning that $B=-A=-1$ Therefore, $$N=\int\frac{dx}{x+1}-\int\frac{dx}{(x+1)^2}$$ $$N=\ln|x+1|+\frac{1}{x+1}+C$$ So, in conclusion, $$A=M+N=\frac{1}{6}(\tan^{-1}(3x))^2+\ln|x+1|+\frac{1}{x+1}+C$$