Evaluate the definite integral ${\int }_0^2\frac{x^2+1}{(x+1{)}^2}dx$ using algebraic manipulation and substitution.

Step-by-step solution

1. Rewrite the integrand by dividing $x^2+1$ by $(x+1{)}^2$. Use the identity $x^2+1=(x+1)(x-1)+2$. Thus $\frac{x^2+1}{(x+1{)}^2}=\frac{x-1}{x+1}+\frac{2}{(x+1{)}^2}$.
2. Integrate the first term: let $t=x+1$, then $x-1=t-2$. $\int \frac{x-1}{x+1}dx=\int (1-\frac{2}{t})dt=t-2\ln |t|$. Replace $t$ with $x+1$: $=x+1-2\ln (x+1)$. The second term integrates to $\int \frac{2}{(x+1{)}^2}dx=-\frac{2}{x+1}$. Add both results: $x+1-2\ln (x+1)-\frac{2}{x+1}$. Apply the limits 0 to 2: At $x=2$: $3-2\ln 3-\frac{2}{3}=\frac{7}{3}-2\ln 3$. At $x=0$: $1-0-2=-1$. Subtract: $(\frac{7}{3}-2\ln 3)-(-1)=\frac{10}{3}-2\ln 3$.

Concept tested: Definite integral, algebraic manipulation

Published 5 September 2026 · How this solution was prepared and checked