How do I solve $\int\frac{dy}{2y+y^2}$?

Question

the integral is:

$$\int\frac{dy}{2y+y^2}$$

So I had:

$$\frac{1}{2}\int\frac{dy}{y(y-1)} = \frac{A}{y} + \frac{B}{y-1}$$

so $$\begin{cases}A=-1 \\ B=1 \end{cases}$$

$$\frac{1}{2}\left(-\int\frac{dy}{y}+\int\frac{dy}{y-2}\right)$$

and the end $$\frac{1}{2}\left(-\ln|y|+\ln|y-2|\right) + C$$

But original answer is:

$$\frac{1}{2}\left(\ln|x-2|-\ln|x|\right)+C$$

What did I miss?

Answer 1

If you hate partial fraction decomposition, you could also try completing the square in the denominator. Doing this yields

\begin{align*} \int \frac{1}{y^2+2y} dx = \int \frac{1}{(y+1)^2 - 1 } dx \end{align*}

Now make the substitution that $\sec(\theta) = y + 1 \ \Rightarrow\ dy = \sec\theta\tan{\theta}d\theta$. Therefore,

\begin{align*} \int \frac{1}{(y+1)^2 - 1 } dx & = \int \frac{\sec\theta\tan\theta}{\sec^{2}\theta - 1} d\theta\\ &=\int \frac{\sec\theta\tan\theta}{\tan^2\theta} d\theta\\ &= \int \csc(\theta) d\theta\\ &= -\ln|\csc\theta + \cot\theta| + C. \end{align*}

Now make the usual backward trig subs for your answer.

Answer 2

This

$\int\frac{dy}{y(y+2)} = \frac{A}{y} + \frac{B}{y+2}$

\begin{cases} Ay+By=0 \iff A=-B\\ 2A=1 \iff A= \frac 1 2 \end{cases}

Then (A,B)=(1/2,-1/2)

The solution is $g(y)= -\frac {1} 2\ln \mid (y+2) \mid+ \frac 1 2 \ln\mid y\mid + K$