Partial Fraction Decomposition??

Question

How do I separate this by partial fraction decomposition?

$$\int\frac1{u(u^2 + 1)}du$$

I've used the normal technique and got to:

$$1 = A(u^2 + 1) + B(u)$$

and $A=1$ if $u=0$ BUT how do I find $B$ now? because I can't make $u^2 +1$ equal to 0.

Answer 1

Patial fractions work when you want to split $$\frac{1}{P(x)Q(x)}$$

into $$\frac{A}{P(X)} + \frac{B}{Q(X)}$$

where $P$ and $Q$ are polynomials.

You don't have that. You have just one polynomial in the denominator.

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Hint:

To calculate the integral, review the derivatives of the basic trigonometric functions and their inverses.

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AFTER YOUR EDIT:

You made a mistake where if the denominator polynomial has degree $n$, then the numerator needs to have degree $n-1$, so in your case, you should set

$$\frac{1}{u(u^2+1)} = \frac Au + \frac{Bu + C}{u^2+1}$$

Now it should be easy to get the solution.

Answer 2

The real partial fraction decomposition of $ \frac{1}{u^2+1}$ is $ \frac{1}{u^2+1}$ , since $u^2+1$ has no real zeroes.

We have $u^2+1=(u-i)(u+i)$. For the complex partial fraction decomposition determine $A$ and $B$ such that

$ \frac{1}{u^2+1}= \frac{A}{u-i}+\frac{B}{u+i}$.

But do not use this decomposition for $\int \frac{1}{u^2+1} du$, since

$\int \frac{1}{u^2+1} du=\arctan (u) +C$

Answer 3

Solution without partial Decomposition::

$$I = \int\frac{1}{u(u^2+1)}du = \int\frac{1}{u^3(1+u^{-2})}du$$

Now Put $1+u^{-2} = t\;,$ Then $\displaystyle -\frac{2}{u^3}du = dt\Rightarrow \frac{1}{u^3}du = -\frac{1}{2}dt$

So $$I = -\frac{1}{2}\int\frac{1}{t}dt = -\frac{1}{2}\ln t+\mathcal{C} = -\frac{1}{2}\ln \left|\frac{1+u^2}{u^2}\right|+\mathcal{C}=\frac{1}{2}\ln \left|\frac{u^2}{1+u^2}\right|+\mathcal{C}$$