How can I integrate $\int \frac{dx}{(1+x^2)(2+x)}$ using partial fraction?

Question

I did this: $$ \frac{1}{(1+x^2)(2+x)} = \frac{A}{1+x^2} + \frac{B}{2+x} \Rightarrow 2A+B+Ax+ Bx^2 = 1 $$ but I do not know what to do next. Every help will be appreciated.

Answer 1

Hint

Your decomposition is wrong, the more genral form is $$ \frac{Ax+C}{1+x^2} + \frac{B}{x+2} $$ and this will give you 3 equations, one for each power of $x: 1, x, x^2$, which you can use to solve for 3 variables $A, B, C$.

Then, $$ \int \frac{dx}{\left(1+x^2\right)(x+2)} = \int \frac{(Ax + C)dx}{1+x^2} + \int \frac{Bdx}{x+2}, $$ both of which are standard forms easy to evaluate.

Answer 2

As @gt69896 said, your setup is incorrect - you must have $3$ coefficients, $A$, $B$, and $C$. Now, once you get the correct numerator, it will be a polynomial in $x$ with coefficients depending on $A$, $B$, and $C$. That polynomial must be equal to $1$ for all $x$.

There are several ways to find $A$, $B$, and $C$. One method, as @gt69896 suggested, is to replace $x$ with three different values to get three equations in $A$, $B$, and $C$. Since $x= -2$ makes $2+ x$ equal to $0$, that would be a good choice. There are no real values that make $x^2+ 1$ equal to $0$ so just choose simple values - $x= 0$ and $x= 1$ would be good.

Another way is to use the fact that if two polynomials, in $x$, are equal for all $x$ then each term must have the same coefficients.

For example, if the equation were $2A+ B+ Ax+ Bx^2= Bx^2+ Ax+ (2A+ B)= 0x^2+ 0x+ 1$, as you give, then you must have $B= 0, A= 0, 2A+ B= 1$. Of course, there are NO values of $A$ and $B$ that will satisfy those three equations (was that your difficulty?) which is why we need "$Ax+ B$" and "$C$" as numerators.