anonymous
  • anonymous
evaluate the integral from x=2 to x=5 of 1/2 ln(x^2 +25) - 1/2 ln(x^2 +4) dx
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
I'm seeing that your integral has something in common: ln(x^2+25) and ln(x^2+4) have the form,\[\ln(x^2+a^2)\] (one a is 5, the other, 2). To make light of the work, I think the integral\[\int\limits_{}{}\ln (x^2+a^2)dx\]need only be considered. You can work out what you need then by applying the result a couple of times (i.e. with different a's.
anonymous
  • anonymous
Start with integration by parts. Let \[u=\ln(x^2+a^2) \rightarrow du=\frac{2x}{x^2+a^2}dx\]and\[dv=1 \rightarrow v=x\]Then\[\int\limits_{}{}\ln (x^2+a^2)dx=x \ln (x^2+a^2)-2\int\limits_{}{}\frac{x^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)-2\int\limits_{}{}\frac{x^2+a^2-a^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)-2\int\limits_{}{}\frac{x^2+a^2}{x^2+a^2}-\frac{a^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)-2\int\limits_{}{}1-\frac{a^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)-2x+2a^2\int\limits_{}{}\frac{dx}{x^2+a^2}\]\[=x \ln (x^2+a^2)-2x+2a \tan^{-1}\frac{x}{a}+c\]
anonymous
  • anonymous
You can obtain the last integral using the substitution u=(x/a) after making the rearrangement,\[\int\limits_{}{}\frac{dx}{x^2+a^2}=\frac{1}{a^2}\int\limits_{}{}\frac{dx}{(\frac{x}{a})^2+1} \rightarrow \frac{1}{a^2}\int\limits_{}{}\frac{a}{1+u^2}du\]\[=\frac{1}{a}\tan^{-1} u+c=\frac{1}{a}\tan^{-1} \frac{x}{a}+c\]

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anonymous
  • anonymous
very complex
anonymous
  • anonymous
\[\frac{1}{2}\int\limits_{2}^{5}\ln (x^2+25)-\ln (x^2+4)dx\]\[\frac{1}{2}\int\limits_{2}^{5}\ln (x^2+25)dx- \frac{1}{2}\int\limits_{2}^{5}\ln (x^2+4)dx\]\[=\frac{1}{2}[x \ln (x^2+25)-2x+10\tan^{-1}\frac{x}{5}\]\[-x \ln (x^2+4)+2x-4\tan^{-1}\frac{x}{2}]^{5}_{2}\]

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