How do you solve the integral $\int \frac{2x+5}{x^2+2x+5} \,dx $ with any method?

Question

I am curious how many ways there are to solve this integral:

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

I have a solution with one method, but I would like to know how you solve the integral with the method you prefer.

If it's possible, could you tell me the country you're come from, please? I'm making an experiment.

Thanks.

Answer 1

Well note that $$\int\frac{2x+5}{x^{2}+2x+5}= \int \frac{2x+2}{x^{2}+2x+5} + \int \frac{3}{x^{2}+2x+5}$$

The first integral is easy. For the second one note that $$x^{2}+2x+5 = (x+1)^{2} + 2^{2}$$ and so put $x+1 = 2\tan(t)$ and try to solve it.

Answer 2

$$I=\int \frac{2x+5}{x^2+2x+5} \,dx$$ $$I=\int \frac{2x+2}{x^2+2x+5} \,dx+3\int \frac{dx}{(x+1)^2+4} $$ substitute $u={x^2+2x+5}$ and $v=x+1$ $$I=\int \frac{du}{u}+3\int \frac{dv}{v^2+4} $$ And $s=v/2$ $$I=\int \frac{du}{u}+\frac 32\int \frac{ds}{s^2+1} $$ And use the $arctan$ for the last integral... $$I=\ln|x^2+2x+5|+\frac 32 \arctan \left({\frac {x+1}2} \right)+C$$

Answer 3

There are many ways like completing the square and doing some substitution ,or splitting the integral first.

You can also express the numerator as a derivative of the denominator plus some constant like so;

$I = \int \frac{2x+5}{x^2+2x+5} \,dx$

let $2x+5 =A(\frac{d}{dx}(x^2+2x+5))+B$

$2x+5 = A(2x+2)+B$

$2x+5 =2Ax+2A+B$

$\therefore A= 1 $ and $B = 3$

$I=\int \frac{2x+2+3}{x^2+2x+5}\,dx$

$I = \int\frac{2x+2}{x^2+2x+5}+\frac{3}{x^2+2x+5}\,dx $

The first integral can be solved using a simple substitution of $u =x^2+2x+5$.

For the second integral ;

$J =\int \frac3{x^2+2x+5}\,dx$

$ = \int \frac3{(x+1)^2+4}\,dx$

$= \frac1{6}\arctan(\frac{x+1}2)+C$