-1
$\begingroup$

I've tried pulling out 1/2 in front of the integral but it became 1 = Ax+B which I wouldn't be able to find the values of.

Is it possible to leave the 2 in the denominator but long divide with (x^2+1)? integral

  • 0
    Note that the first step in partial fraction decomposition is to eliminate any improper fraction. But once you eliminate the improper fraction in this example the remaining fraction is a partial fraction. So then you are done with the decomposition. – 2017-02-13
  • 0
    @JohnWaylandBales Awesome, thought I was supposed to do more. – 2017-02-13

2 Answers 2

0

Note that $$\int \frac{x^2}{2(x^2+1)} \; \mathrm{d}x=\frac{1}{2} \int \frac{x^2+1-1}{x^2+1} \; \mathrm{d}x=\frac{1}{2}\int \left( 1-\frac{1}{x^2+1} \right) \; \mathrm{d}x$$

Now substitute $x=\tan u$. Or, if you memorized some tables, recall the fact that $$\int \frac{1}{x^2+1} \;\mathrm{d}x=\arctan x$$

  • 0
    My homework explicitly tells me to use partial fraction decomposition, would the long division fulfill that and I can just finish out the problem? – 2017-02-13
  • 0
    @cout Using partial fraction decomposition? Long division, I don't think, would do it. Do you know Euler's Formula? – 2017-02-13
  • 0
    I don't, would it help with this problem? I'm not into series yet in my class. – 2017-02-13
  • 0
    @cout Do you know then, that $$\arctan(x) = \frac{-1}{2i}\ln(x+i)+\frac{1}{2i}\ln(x-i) + c$$ That is necessary to do this using partial fraction decomposition. – 2017-02-13
0

We have $$I = \frac {1}{2} \int \frac {x^2}{1+x^2} \mathrm {d}x $$ $$ = \frac {1}{2} \int \frac {x^2+1-1}{1+x^2} \mathrm {d}x $$ $$=\frac {1}{2}[ \int \frac {x^2+1}{x^2+1} \mathrm {d}x - \int \frac {1}{x^2+1} \mathrm{d}x] $$ $$=\frac {1}{2}[\int 1 \mathrm {d}x - \int \frac {1}{x^2+1} \mathrm {d}x] $$

Hope you can take it from here.