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How to integrate $\displaystyle \int \frac{2x^2+x}{(x+1)(x^2+1)}dx$? I Tried using partial fractions but i got lost, thanks.

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Partial fractions are the way to go. The fraction is already reduced, and the denominator is fully factored over the reals, so your setup is

$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)(x+1)}{(x+1)(x^2+1)}\;,$$

and you must find $A,B$, and $C$ so that $$2x^2+x=A(x^2+1)+(Bx+C)(x+1)=(A+B)x^2+(B+C)x+(A+C)\;.$$

Equating coefficients of powers of $x$ yields the system

$$\left\{\begin{align*} &A+B=2\\ &B+C=1\\ &A+C=0\;, \end{align*}\right.$$

which is easily solved: $A=\frac12,B=\frac32$, and $C=-\frac12$. Thus,

$$\frac{2x^2+x}{(x+1)(x^2+1)}=\frac1{2(x+1)}+\frac{3x-1}{2(x^2+1)}\;,$$ and

$$\int\frac{2x^2+x}{(x+1)(x^2+1)}dx=\frac12\int\frac1{x+1}dx+\frac12\int\frac{3x-1}{x^2+1}dx\;.$$

You shouldn’t have any trouble with $\int\frac1{x+1}dx$. The other term is most easily handled by splitting it:

$$\int\frac{3x-1}{x^2+1}dx=3\int\frac{x}{x^2+1}dx-\int\frac1{x^2+1}dx\;,$$

where the first integral succumbs to a $u$-substitution, and the second is one that you should know (or at least be able to work by a trig substitution).

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  • $\begingroup$ @Marvis: You’re right, of course. I’m fixing it now. $\endgroup$ – Brian M. Scott Oct 19 '12 at 5:16
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    $\begingroup$ There is a trick known as the "cover-up rule" which gets you the constants which go with linear factors - in the expression for $2x^2+x$ set $x=-1$ which immediately gives you an expression for $A$. In this case you could deal with the quadratic factor and get $B$ and $C$ by setting $x=\pm i$. It is called the cover-up rule because it is computable directly from the original expression for partial fractions by computing the LHS after covering up a factor in the denominator and evaluating what remains. $\endgroup$ – Mark Bennet Oct 19 '12 at 7:43
  • $\begingroup$ @MarkBennet: I usually choose this "rule" for finding constants. It often works good. $\endgroup$ – mrs Oct 19 '12 at 8:38
  • $\begingroup$ @Mark: Interesting: I’d never heard it given a name. It’s a nice shortcut, but I definitely think of it as an extra. $\endgroup$ – Brian M. Scott Oct 19 '12 at 12:55
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The key is to write $(2x^2 + x)$ as $A(x^2+1) + (Bx+C)(x+1)$ $$(2x^2 + x) = A(x^2+1) + (Bx+C)(x+1) = (A+B)x^2 + (B+C)x + (A+C)$$ This gives us $A+B = 2$, $B+C = 1$ and $A+C = 0$ i.e. $A+B = 2$ and $B-A = 1$.

$$A = \dfrac12, B = \dfrac32, C = -\dfrac12$$

Hence, $$\dfrac{2x^2+x}{(x+1)(x^2+1)} = \dfrac1{2(x+1)} + \dfrac{3x-1}{2(x^2+1)} = \dfrac1{2(x+1)} + \dfrac34 \dfrac{2x}{x^2+1} - \dfrac12 \dfrac1{x^2+1}$$

Now you should be able to integrate it.

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