Question

How do you find `int (x+4)/(x^2 + 2x + 5) dx` using partial fractions?

Answer 1

`\color{red}{\int\frac{x+4}{x^2+2x+5}\ dx}=\color{blue}{1/2\ln|x^2+2x+5|+\frac{3}{2} \tan^{-1}(\frac{x+1}{2})+C`

Explanation:

`\int\frac{x+4}{x^2+2x+5}\ dx`

`=\int \frac{1/2(2x+2)+3}{x^2+2x+5}\ dx`

`=1/2\int \frac{(2x+2)dx}{x^2+2x+5}+\int \frac{3}{x^2+2x+5}\ dx`

`=1/2\int \frac{d(x^2+2x+5)}{x^2+2x+5}+3\int \frac{1}{(x+1)^2+2^2}\ dx`

`=1/2\ln|x^2+2x+5|+3\int \frac{d(x+1)}{(x+1)^2+2^2}`

Using standard formula `\int \frac{1}{t^2+a^2}dt=1/a\tan^{-1}(t/a)`,

`=1/2\ln|x^2+2x+5|+3\frac{1}{2} \tan^{-1}(\frac{x+1}{2})+C`

`=1/2\ln|x^2+2x+5|+\frac{3}{2} \tan^{-1}(\frac{x+1}{2})+C`