Question

Solve $\int \frac{2+x+x^2}{x^2\left(2+x\right)}+\frac{2x-1}{{\left(x+1\right)}^2}dx$

• A. $\ln \left|2-x\right|+\frac{3}{\left(x+1\right)}+2\ln \left(x+1\right)-\frac{1}{x}+C$
• B. $\ln \left|2+x\right|+\frac{3}{\left(x+1\right)}+2\ln \left(x+1\right)-\frac{1}{x}+C$
• C. $\ln \left|2+x\right|+\frac{3}{\left(x-1\right)}+2\ln \left(x+1\right)-\frac{1}{x}+C$
• D. None of these

Answer 1

The correct option is B $\ln \left|2+x\right|+\frac{3}{\left(x+1\right)}+2\ln \left(x+1\right)-\frac{1}{x}+C$

$\begin{array}{c}\int \frac{2+x+x^2}{x^2\left(2+x\right)}+\frac{2x-1}{{\left(x+1\right)}^2}dx\\ =\int \frac{2+x}{\left(2+x\right)x^2}+\frac{x^2}{x^2\left(2+x\right)}+\frac{2x+2}{{\left(x+1\right)}^2}-\frac{3}{{\left(x+1\right)}^2}dx\\ =\int \frac{1}{x^2}+\frac{1}{2+x}+\frac{2\left(x+1\right)}{{\left(x+1\right)}^2}-\frac{3}{{\left(x+1\right)}^2}dx\\ =-\frac{1}{x}+\ln \left|2+x\right|+\frac{3}{\left(x+1\right)}+\int \frac{2\left(x+1\right)}{{\left(x+1\right)}^2}dx\end{array}$

Let

$\begin{array}{c}{\left(x+1\right)}^2=t\\ 2\left(x+1\right)dx=dt\\ =-\frac{1}{x}+\ln \left|2+x\right|+\frac{3}{\left(x+1\right)}+\int \frac{2}{t}dt\\ =-\frac{1}{x}+\ln \left|2+x\right|+\frac{3}{\left(x+1\right)}+\ln {\left(x+1\right)}^2+C\\ =\ln \left|2+x\right|+\frac{3}{\left(x+1\right)}+2\ln \left(x+1\right)-\frac{1}{x}+C\end{array}$