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?
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2 Answers
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$\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 $\displaystyle \int \log(t) dt = t (\log(t) -1) + c$
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$\begingroup$ Yes, thats not the problem. But what do I do with the sides($1+x^2$ and $1-x^2$)? $\endgroup$ Feb 22, 2013 at 13:00
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1$\begingroup$ same as you do with limits of numerical values, of course, the result will be function of $x$ $\endgroup$– S LFeb 22, 2013 at 13:04
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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.
