What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$?

Question

I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?

Answer 1

First of all: you have to be carefull with your notation because you mean:

$$\int\left(\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\right)\space\text{d}x$$

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And Partial fractions is the most easy way:

$$\int\frac{x-1}{(x+3)(x^2+1)}\space\text{d}x=$$ $$\int\left(\frac{2x-1}{5(x^2+1)}-\frac{2}{5(x+3)}\right)\space\text{d}x=$$ $$\frac{1}{5}\int\frac{2x-1}{x^2+1}\space\text{d}x-\frac{2}{5}\int\frac{1}{x+3}\space\text{d}x=$$ $$\frac{1}{5}\int\left(\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\space\text{d}x-\frac{2}{5}\int\frac{1}{x+3}\space\text{d}x$$

Answer 2

$\int \frac{x-1}{(x+3)(x^2+1)} dx= \int\frac{A}{x+3} + \frac{B(2x) + C}{x^2+1}\,dx $ and

$ x-1= A(x^{2}+1)+ (B(2x)+C)(x+3)$ which is simple.

Answer 3

Notice, for $(Bx+C)$ part, you should use separation as follows $$\int \frac{x-1}{(x+3)(x^2+1)}\ dx=\int \frac{-2}{5(x+3)}+\frac{2x-1}{5(x^2+1)}\ dx$$ $$=\frac{1}{5}\int \left(-\frac{2}{x+3}+\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\ dx$$ $$=\frac{1}{5} \left(-2\int \frac{1}{x+3}\ dx+\int \frac{d(x^2)}{x^2+1}-\int\frac{1}{x^2+1}\ dx\right)$$

Answer 4

Here is a slightly different way. We start with $$ \frac{x - 1}{(x+3)(x^2+1)} = \frac{A}{x+3} + \frac{B}{x+i} + \frac{B^*}{x-i}, \qquad (1) $$ where $B^*$ is the complex conjugate of $B$. The coefficient for $(x-i)^{-1}$ must be $B^*$: since the left-hand side is real, the right-hand side is real too, and hence is invariant under complex conjugation.

Multiplying both sides by $x+3$ and taking the limit of $x \rightarrow -3$, we get $$ A = \frac{-3-1}{(-3)^2 + 1} = -\frac{2}{5}. \qquad (2) $$

Similarly, multiplying both sides by $x + i$ and taking the limit of $x \rightarrow -i$ yields $$ B = \frac{-i - 1}{(-i + 3)(-i - i)} = \frac{1}{5} - \frac{i}{10}. \qquad (2) $$

Now the right-hand side of (1) is easy to integrate $$ \begin{aligned} \int \frac{x - 1}{(x+3)(x^2+1)} dx &= A\log|x+3| + B\log(x+i) + B^*\log(x-i) + C\\ &= A \log|x+3| + (\Re B) \log(x+i)(x-i) + (\Im B) \, i\log\frac{x+i}{x-i} + C \\ &= A \log|x+3| + (\Re B) \log(x^2+1) -2 (\Im B) \arctan \frac{1}{x} + C \\ &= -\frac{2}{5} \log|x+3| + \frac{1}{5} \log(x^2+1) +\frac{1}{5} \arctan \frac 1 x + C. \end{aligned} $$

The third line needs some explanation. Since $$ \begin{aligned} \log \frac{x + i}{\sqrt{x^2+1}} &= i \arctan\frac{1}{x}, \\ \log \frac{x - i}{\sqrt{x^2+1}} &= -i \arctan\frac{1}{x}, \end{aligned} $$ the difference $$ \log \frac{x + i}{x - i} = 2 i \arctan\frac{1}{x}. $$