Integral of x^2/(x+2) Solution | Evaluate the Integral

  • Thread starter Thread starter professordad
  • Start date Start date
  • Tags Tags
    Integral
professordad
Messages
18
Reaction score
0

Homework Statement



Evaluate the integral:
\int \frac{x^2}{x + 2} dx

Homework Equations



There aren't any relevant equations...

The Attempt at a Solution



In these type of problems you have to set u = \text{something} so I tried setting u = x^2, but then \text{du} = 2x\text{dx} and you can't substitute anything. And if u = x + 2 then \text{du} = 1 and that's useless.

Also, this problem is from the Swokowski calculus textbook (school starts in two days for me so I'm doing self study :smile: )
 
Last edited:
Physics news on Phys.org
Before you do anything, do polynomial division (degree of top >= degree of bottom). Once this is done, it should be trivial to take the integral.
 
Try doing some algebra first. You can rewrite that. Polynomial long division.
 
In you second substitution, where u = x+2, du = dx not du = 1

If you carry thru with this substitution properly, you can find the desired integral.
 
Wow, I can't believe I missed that. Here's what I have from there:

Clearly \frac{x^2}{x + 2} = x - 2 + \frac{4}{x + 2}. So integrating that we have \frac{x^2}{2} - 2x + 4\ln{|x+2|} + C where C is a constant.

Thanks for your help guys! I appreciated it :smile:

Also, to SteamKing, yes, I made a mistake while typing it up.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top