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.


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


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


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


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).

  • $\begingroup$ @Marvis: You’re right, of course. I’m fixing it now. $\endgroup$ – Brian M. Scott Oct 19 '12 at 5:16
  • 1
    $\begingroup$ 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. $\endgroup$ – Mikasa 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

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.


From $\frac{2x^2+ x}{(x+ 1)(x^2+ 1)}= \frac{A}{x+ 1}+ \frac{Bx+ C}{x^2+ 1}$ you can add the fractions on the right, then compare coefficients of "like terms". But we can also use the fact that this is to be true for all x. Setting x to any three values gives three equations to solve for A, B, and C.

First multiplying both sides by $x+1$ and $x^2+ 1$ simplifies to $2x^2+ x= A(x^2+ 1)+ (Bx+ C)(x+ 1).

Setting x= -1 immediately gives $2(1)+ (-1)= 1= A(2)$ so $A= \frac{1}{2}$.

Setting x= 0 gives $0= A+ C= \frac{1}{2}+ C$ so $C= -\frac{1}{2}$.

Setting x= 1 gives $2+ 1= 3= 2A+ 2(B+ C)= 1+ 2B- 1$ so $B= \frac{3}{2}$.


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