# Lengthy integration problem

$$\int\frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx$$

I managed to solve the problem using partial fraction decomposition.

But that approach is pretty long as it creates five variables. Is there any other shorter method to solve this problem(other than partial fractions)?

I also tried trigonometric substitution and creating the derivative of the denominator in the numerator... but it becomes even longer. Thanks in advance!!

Just to clarify: (My partial fraction decomposition)

$$\frac{x^3+3x+2}{(x^2+1)^2(x+1)}= \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2} + \frac{E}{x+1}$$

• What do you mean when you state that “it creates five variables”? The answer is $\frac{1}{4} \left(\frac{4 x}{x^2+1}+\log \left(x^2+1\right)-2 \log (x+1)+6 \arctan(x)\right)+C$. Commented Jul 19, 2017 at 5:18
• I believe its in reference to the linear system of equations made during partial fraction decomposition @JoséCarlosSantos Commented Jul 19, 2017 at 5:20
• @JonathanDavidson Me too. But I would like to have the OP point of view. Commented Jul 19, 2017 at 5:21

HINT:

$$x^3+3x+2=x^3+x+2x+2=x(x^2+1)+2(x+1)$$

$$\dfrac{x^3+3x+2}{(x^2+1)^2(x+1)}=\dfrac x{(x^2+1)(x+1)}+\dfrac2{(x^2+1)^2}$$

For the second set $x=\tan y$

For the first

Method$\#1:$

write the numerator as $$x\cdot\dfrac{(x^2+1)-(x^2-1)}2$$

Method$\#2:$

set $x=\tan y$ to find $$\int\dfrac{\sin y}{\sin y+\cos y}dx$$

Now express $\sin y=A(\sin y+\cos y)+B\cdot\dfrac{d(\sin y+\cos y)}{dy}$

Can you derive the values of the arbitrary constants $A,B?$

There are much better ways to find the coefficients in partial fractions than solving five equations in five variables.

$$\frac{x^3 + 3 x + 2}{(x^2+1)^2 (x+1)} = \frac{Q(x)}{(x^2+1)^2} + \frac{E}{x+1}$$ multiply both sides by $x+1$ and substitute $x=-1$. We get $$\frac{-2}{2^2} = 0 + E$$ so $E = -1/2$, and \eqalign{\frac{Q(x)}{(x^2+1)^2} &= \frac{x^3 + 3 x + 2}{(x^2+1)^2 (x+1)} + \frac{1/2}{x+1}\cr &= \frac{x^2 + 3 x + 2 + (1/2)(x^2+1)^2)}{(x^2+1)^2(x+1)}\cr &= \frac{x^3 + x^2 + x + 5}{2(x^2 + 1)^2} } Now you want this in the form $$\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$$ Multiply by $(x^2 + 1)^2$ and substitute $x=i$ (yes, $\sqrt{-1}$). We get $$2 = C i + D$$ But since we want $C$ and $D$ to be real, we must have $C = 0$, $D = 2$. This leaves \eqalign{\frac{Ax+B}{x^2+1} &= \frac{x^3 + x^2 + x + 5}{2(x^2 + 1)^2} - \frac{1}{(x^2+1)^2} \cr &= \frac{x + 1}{2(x^2 + 1)}} and we're done: $A = 1/2$, $B = 1/2$, $C = 0$, $D = 2$, $E = -1/2$.