Mathematics - Indefinite Integration Question with Solution | TestHub

Let $I=\int \frac{\ln x}{(x+1{)}^2}dx$. We use integration by parts, $\int udv=uv-\int vdu$. Let $u=\ln x$ and $dv=\frac{1}{(x+1{)}^2}dx$. Then $du=\frac{1}{x}dx$ and $v=\int \frac{1}{(x+1{)}^2}dx=-\frac{1}{x+1}$. Applying the formula:$I=(\ln x)\left(-\frac{1}{x+1}\right)-\int \left(-\frac{1}{x+1}\right)\frac{1}{x}dx$ $I=-\frac{\ln x}{x+1}+\int \frac{1}{x(x+1)}dx$. For the remaining integral, we use partial fractions: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}$. So, $\int \frac{1}{x(x+1)}dx=\int \left(\frac{1}{x}-\frac{1}{x+1}\right)dx=\ln |x|-\ln |x+1|+C^{\prime }=\ln \left|\frac{x}{x+1}\right|+C^{\prime }$. Combining these results, we get $I=-\frac{\ln x}{x+1}+\ln \left|\frac{x}{x+1}\right|+C$.