$\int \ln \left(\frac{1}{x}\right) d x=-x \ln x+x+C$ .

Name: $\int \ln \left(\frac{1}{x}\right) d x=-x \ln x+x+C$ .
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Question

Quizplus · Accepted Answer

The integral $\int \ln \left(\frac{1}{x}\right) d x$ simplifies to $\int -\ln(x) dx$ because $\ln \left(\frac{1}{x}\right) = -\ln(x)$. Using integration by parts, where $u = -\ln(x)$ and $dv = dx$, gives $uv - \int v du = -x \ln(x) + \int dx = -x \ln(x) + x + C$, which matches the given expression.

∫ln⁡(1x)dx=−xln⁡x+x+C\int \ln \left(\frac{1}{x}\right) d x=-x \ln x+x+C∫ln(x1​)dx=−xlnx+x+C

$\int \ln \left(\frac{1}{x}\right) d x=-x \ln x+x+C$