Integration by Parts Math Example 5

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Example 5

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Find โˆซlnโกxโ€‰dx\displaystyle\int \ln x\,dx.

Solution

  1. 1
    Write as โˆซlnโกxโ‹…1โ€‰dx\int \ln x \cdot 1\,dx. Let u=lnโกxu = \ln x, dv=dxdv = dx; du=dxxdu = \frac{dx}{x}, v=xv = x.
  2. 2
    xlnโกxโˆ’โˆซxโ‹…1xโ€‰dx=xlnโกxโˆ’x+Cx\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C.

Answer

xlnโกxโˆ’x+Cx\ln x - x + C
Writing lnโกx=lnโกxโ‹…1\ln x = \ln x \cdot 1 sets up IBP with dv=dxdv = dx. This is the standard trick for integrating isolated logarithms.

About Integration by Parts

An integration technique based on the product rule: โˆซuโ€‰dv=uvโˆ’โˆซvโ€‰du\int u\,dv = uv - \int v\,du. Used when the integrand is a product of two functions.

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