$\int \frac{x}{x^{2}-9} d x=\frac{1}{2} \ln \left|x^{2}-9\right|+C$ .

Name: $\int \frac{x}{x^{2}-9} d x=\frac{1}{2} \ln \left|x^{2}-9\right|+C$ .
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The integral $\int \frac{x}{x^{2}-9} dx$ can be solved by using the substitution $u = x^2 - 9$, which simplifies the integral to $\frac{1}{2} \ln|u| + C$, or $\frac{1}{2} \ln|x^2 - 9| + C$, because the derivative of $x^2 - 9$ is $2x$, and the $x$ in the numerator cancels out the $2$ in the denominator after adjusting for the differential $du = 2x dx$.

∫xx2−9dx=12ln⁡∣x2−9∣+C\int \frac{x}{x^{2}-9} d x=\frac{1}{2} \ln \left|x^{2}-9\right|+C∫x2−9x​dx=21​ln​x2−9​+C

$\int \frac{x}{x^{2}-9} d x=\frac{1}{2} \ln \left|x^{2}-9\right|+C$