Question

$\int \frac{x+2}{2x^2+6x+5}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{x+2}{2x^2+6x+5}\right)dx \\ x+2=\mathrm{A}\frac{d}{dx}\left(2x^2+6x+5\right)+\mathrm{B} \\ x+2=\mathrm{A}\left(4x+6\right)+\mathrm{B} \\ x+2=\left(4\mathrm{A}\right)x+6\mathrm{A}+\mathrm{B} \end{gathered}$$

Comparing the Coefficients of like powers of x

$$\begin{gathered} 4\mathrm{A}=1 \\ \mathrm{A}=\frac{1}{4} \\ 6\mathrm{A}+\mathrm{B}=2 \\ 6\times \frac{1}{4}+\mathrm{B}=2 \\ \mathrm{B}=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \therefore \int \left(\frac{x+2}{2x^2+6x+5}\right)dx \\ =\int \left[\frac{{\displaystyle \frac{1}{4}}\left(4x+6\right)+{\displaystyle \frac{1}{2}}}{2x^2+6x+5}\right]dx \\ =\frac{1}{4}\int \frac{\left(4x+6\right)}{2x^2+6x+5}dx+\frac{1}{2}\int \frac{1}{2x^2+6x+5}dx \\ =\frac{1}{4}\int \frac{\left(4x+6\right)}{2x^2+6x+5}dx+\frac{1}{4}\int \frac{dx}{x^2+3x+{\displaystyle \frac{5}{2}}} \\ =\frac{1}{4}\int \frac{\left(4x+6\right)}{2x^2+6x+5}dx+\frac{1}{4}\int \frac{dx}{x^2+3x+{\left({\displaystyle \frac{3}{2}}\right)}^2-{\left({\displaystyle \frac{3}{2}}\right)}^2+{\displaystyle \frac{5}{2}}} \\ =\frac{1}{4}\int \frac{\left(4x+6\right)}{2x^2+6x+5}dx+\frac{1}{4}\int \frac{dx}{{\left(x+{\displaystyle \frac{3}{2}}\right)}^2-{\displaystyle \frac{9}{4}}+{\displaystyle \frac{5}{2}}} \\ =\frac{1}{4}\int \frac{\left(4x+6\right)}{2x^2+6x+5}dx+\frac{1}{4}\int \frac{dx}{{\left(x+{\displaystyle \frac{3}{2}}\right)}^2{\displaystyle +\frac{1}{4}}} \\ =\frac{1}{4}\int \frac{\left(4x+6\right)}{2x^2+6x+5}dx+\frac{1}{4}\int \frac{dx}{{\left(x+{\displaystyle \frac{3}{2}}\right)}^2+{\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =\frac{1}{4}\log \left|2x^2+6x+5\right|+\frac{1}{4}\times 2{\tan }^{-1}\left(\frac{x+{\displaystyle \frac{3}{2}}}{{\displaystyle \frac{1}{2}}}\right)+C \\ =\frac{1}{4}\log \left|2x^2+6x+5\right|+\frac{1}{2}{\tan }^{-1}\left(2x+3\right)+C \end{gathered}$$