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anonymous
 5 years ago
evaluate the integral from x=2 to x=5 of 1/2 ln(x^2 +25)  1/2 ln(x^2 +4) dx
anonymous
 5 years ago
evaluate the integral from x=2 to x=5 of 1/2 ln(x^2 +25)  1/2 ln(x^2 +4) dx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'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
 5 years ago
Best ResponseYou've already chosen the best response.0Start 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\[\int\limits_{}{}\ln (x^2+a^2)dx=x \ln (x^2+a^2)2\int\limits_{}{}\frac{x^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)2\int\limits_{}{}\frac{x^2+a^2a^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)2\int\limits_{}{}\frac{x^2+a^2}{x^2+a^2}\frac{a^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)2\int\limits_{}{}1\frac{a^2}{x^2+a^2}dx\]\[=x \ln (x^2+a^2)2x+2a^2\int\limits_{}{}\frac{dx}{x^2+a^2}\]\[=x \ln (x^2+a^2)2x+2a \tan^{1}\frac{x}{a}+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can obtain the last integral using the substitution u=(x/a) after making the rearrangement,\[\int\limits_{}{}\frac{dx}{x^2+a^2}=\frac{1}{a^2}\int\limits_{}{}\frac{dx}{(\frac{x}{a})^2+1} \rightarrow \frac{1}{a^2}\int\limits_{}{}\frac{a}{1+u^2}du\]\[=\frac{1}{a}\tan^{1} u+c=\frac{1}{a}\tan^{1} \frac{x}{a}+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{2}\int\limits_{2}^{5}\ln (x^2+25)\ln (x^2+4)dx\]\[\frac{1}{2}\int\limits_{2}^{5}\ln (x^2+25)dx \frac{1}{2}\int\limits_{2}^{5}\ln (x^2+4)dx\]\[=\frac{1}{2}[x \ln (x^2+25)2x+10\tan^{1}\frac{x}{5}\]\[x \ln (x^2+4)+2x4\tan^{1}\frac{x}{2}]^{5}_{2}\]
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