Have a look at this.
Have a look at this.
I get $\dfrac{1}{25-x^2}=\dfrac{1}{10} \left(\dfrac{1}{x+5}-\dfrac{1}{x-5} \right)$
which integrates to
$\dfrac{1}{10}\ln\left(\left|\dfrac{x+5}{x-5}\right|\right)$ and evaluating
$\dfrac{1}{10}\ln\left(\left|\dfrac{x+5}{x-5}\right|\right) \left|^4_0\right.=\\ \\
\dfrac{1}{10}\ln\left(\left|\dfrac{4+5}{4-5}\right|\right)-\dfrac{1}{10}\ln\left(\left|\dfrac{5}{-5}\right|\right)=\\ \\
\dfrac{1}{10}\ln(9)=\\ \\
\dfrac{1}{10}\ln(3^2)=
\dfrac{1}{5}\ln(3)$
you should have been able to figure this out
$\displaystyle \int \dfrac{dx}{a^2-x^2} = \ln\left|\dfrac{a+x}{a-x} \right| + C$
You flipped the fraction on the RHS.
Your answer should be:
$\dfrac{1}{10}\ln\left(\dfrac{5+4}{5-4}\right) - \dfrac{1}{10}\ln\left(\dfrac{5+0}{5-0}\right) = \dfrac{1}{10}\ln 9$
(Since all values are positive, I am ignoring the absolute value symbols).
Next, using logarithm rules:
$\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\cdot \dfrac{1}{2} \ln 9 = \dfrac{1}{5} \ln 9^{1/2} = \dfrac{1}{5}\ln 3$
Romsek beat me to it.
Thanks everyone.
However,where did 1/5 come from?
How does (1/10)*ln 9 become (1/5)(1/2)ln9, which then becomes (1/5)*ln(sqrt{9}) leading to the answer (1/5)*ln 3?
What led you to decide that 1/10 should be broken into two fractions (1/5)(1/2)?
The 1/2 infront of ln becomes the exponent of 9, which is the same as the square root of 9.
No, the OP's answer was wrong:
$\dfrac{1}{10}\ln\left|\dfrac{1}{9}\right| = \dfrac{1}{10}\ln 9^{-1} = -\dfrac{1}{10}\ln 3^2 = -\dfrac{1}{5}\ln 3 \neq \dfrac{1}{5} \ln 3$
nycmath: You used the wrong integral, as I pointed out in my post. That is why you got the wrong answer.
As for why $\dfrac{1}{10}\ln 9 = \dfrac{1}{5}\ln 3$, I broke up the fraction, but that is not needed. You can also do the following:
$\dfrac{1}{10}\ln 9 = \dfrac{1}{10}\ln 3^2 = \dfrac{2}{10}\ln 3 = \dfrac{1}{5}\ln 3$