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

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Answer 1 · 2018-07-18T15:50:21

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

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

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

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

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

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

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

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

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

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