How to integrate a Rational Function $\frac {P(x)}{ Q(x)}$, using Integration By Parts?

Question

If I was given a function with the same degree on both sides (in this case 1)

$$\int \frac{x}{2 \sqrt{x+2}} dx$$

How would I Integrate this using Integration By Parts?

I thought it would be like: $$\frac{1}{2} \int x(x+2)^{-1/2} dx$$ and $u$ would be $x$, $dv$ would be $x(x+2)^{1/2} dx$. But after some trial and error I found that its not.

EDIT: Can a Mod correct the title with $P(x) = Q(x)$

Answer 1

The trick is to "kill" the factor "killable" by derivation and to integrate the factor which is easily integrable.

Considering your exercise, let's define $f(x)=x$ and $g(x)=(x+2)^{-1/2}$. Then you can "kill" $f(x)$ by derivation, because $f'(x)=1$, and you can easily integrate $g(x)$, because $$\int (x+2)^{-1/2} dx= \frac{(x+2)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+c=2\sqrt{x+2}+c$$

Then

$$\int \frac{x}{2 \sqrt{x+2}} dx = \frac{1}{2} \int x(x+2)^{-1/2} dx = \frac{1}{2} \left[ x \left(2\sqrt{x+2} \right) - \int 2\sqrt{x+2} dx \right]$$

And you are done, since $$\int 2\sqrt{x+2} dx = 2\frac{(x+2)^{\frac{1}{2}+1}}{\frac{1}{2}+1} +c = \frac{4}{3}(x+2)^{\frac{3}{2}}+c$$

Then the final solution is

$$x\sqrt{x+2}-\frac{2}{3}(x+2)\sqrt{x+2} +c= \sqrt{x+2} \left( x-\frac{2}{3}(x+2) \right) +c=\frac{1}{3} \sqrt{x+2}(x-4) +c$$

Answer 2

My choice would be to use u-substitution, with $u = x + 2$.

But iff you want to (or need to) use integration by parts, your choices were almost correct:

Put $u = x+2 \implies du = dx$, but you're off on $dv$. Put $$dv = \frac 12 (x + 2)^{-1/2} dx \implies v = \sqrt{x+2}$$

Then you should obtain $$\int \dfrac x{2\sqrt{x + 2}}\,dx = x\sqrt{x+2} - \int \sqrt{x + 2} dx$$

Can you take it from here?

Answer 3

$$\begin{aligned} \ \int \frac{x}{2\sqrt{x+2}} dx &= x\sqrt{x+2} - \int \sqrt{x+2} dx \\ \ &= x\sqrt{x+2} -\frac{2}{3}(x+2)^{3/2} \\ \ &= \sqrt{x+2}\left( x-\frac{2}{3}[x+2]\right) \\ \ &= \sqrt{x+2}\left(\frac{1}{3}x - \frac{4}{3}\right) \\ \ &= \frac{1}{3}(x-4)\sqrt{x+2} \\ \end{aligned} $$

I probably would have stopped (well, actually I did stop the first time round) at the second line, but maybe the final form is a little more informative (tells us the zeroes, etc.).