When can I use the natural log to help solve an integral?
Why is it okay to do this: $\int \frac{1}{x-2}dx = \ln(x-2)$
but not this: $\int \frac{1}{1-x^2}dx = \ln(1-x^2)$
$\endgroup$ 04 Answers
$\begingroup$Because the derivative of $\ln(f(x))$ is not $\frac{1}{f(x)}$ for all differentiable function $f$, even if it is true for $f(x)=x-a$ where $a$ is a constant.
The derivative of $\ln(f(x))$ is $\frac{f^\prime(x)}{f(x)}$ applying the chain rule.
$\endgroup$ $\begingroup$The chain rule is the difference. Note that $\int\frac{du}{u}=\ln|u|$. So, you must have a fraction of the form $u$ on the bottom and the derivative of $u$ on the top. For your second example, $u=1-x^2$, but $du=-2xdx$ is not the numerator.
$\endgroup$ $\begingroup$The first integral can be solved by linear substitution $z = x - 2$.
The second one is non-linear and does not work. Instead you have $-2 \, \int x/(1-x^2) = \mathrm{ln}(1-x^2) + C$. See .
$\endgroup$ $\begingroup$Hint: Try differentiating your second expression on both sides, utlizing FTC.
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