Im trying to resolve the next definite integral: $$\int_{1-x^2}^{1+x^2}{\ln(t^2)\ dt}$$ Im not sure if I can use the Barrow's theorem, I think I have to use the fundamental theorem of integral calculus, but im not sure. How can I solve it?


2 Answers 2


$\displaystyle \log t^2 = 2 \log t, $ so $ \displaystyle \int_{1-x^2}^{1+x^2}{\ln(t^2)·dt} = \int_{1-x^2}^{1+x^2}{2\ln(t)·dt}$ and $\displaystyle \int \log(t) dt = t (\log(t) -1) + c$

  • $\begingroup$ Yes, thats not the problem. But what do I do with the sides($1+x^2$ and $1-x^2$)? $\endgroup$
    – Alejandro
    Feb 22, 2013 at 13:00
  • 1
    $\begingroup$ same as you do with limits of numerical values, of course, the result will be function of $x$ $\endgroup$
    – S L
    Feb 22, 2013 at 13:04
  • $\begingroup$ + Good to note the OP that. $\endgroup$
    – Mikasa
    Feb 23, 2013 at 3:18

If you set $$f(x)=\int_{1-x^2}^{1+x^2}{\ln(t^2)dt}$$ then according to F.T. we get $$f'(x)=4x\ln(1-x^4)$$ You can use the integral by parts firstly to solve the above OE. It takes time to be evaluated so I personally prefer the @experimentX's point of view.


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