## 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

1. 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.

2. 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$

3. 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$

4. anonymous

very complex

5. 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}$