One is a logarithm

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If we are only able to integrate one of the functions, then we simply need to select that as dv/dx.

If we can integrate both functions then a good tip is to select the one you should differentiate as the function that in some way simplifies the next step by for example lowering the grade.

If one is a logarithm we may choose it,  since a logarithm will turn to a reciprocal.

Example:

\int {x\ln x\;dx}  =  

\left| {\begin{array}{*{20}{c}}  {u = \ln x} \\   \begin{gathered}   \frac{{dv}}{{dx}} = x \hfill \\   \frac{{du}}{{dx}} = \frac{1}{x} \hfill \\   v = \frac{{{x^2}}}{2} \hfill \\ \end{gathered} \end{array}} \right.

\begin{gathered}   = \frac{{{x^2}}}{2}\ln x - \int {\frac{1}{x}}  \cdot \frac{{{x^2}}}{2}dx = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\int x \,dx \hfill \\    = \frac{{{x^2}}}{2}\ln x - \frac{{{x^2}}}{4} + C \hfill \\ \end{gathered}  

Another example

I = \int {\frac{{\ln x\;}}{x}dx}  ,  x>0

\left| {\begin{array}{*{20}{c}}    {u = \ln x} \\   \begin{gathered}   \frac{{dv}} {{dx}} = \frac{1}{x} \hfill \\   \frac{{du}}{{dx}} = \frac{1}{x} \hfill \\   v = \ln x \hfill \\ \end{gathered} \end{array}} \right.

 = {(\ln x)^2} - \int {\frac{1}{x}\ln x\;dx = } {(\ln x)^2} - I + {C_1}

Here we can see that the integral has turned into itself, but with the opposite sign. From this we get

I = \tfrac{1}{2}{(\ln x)^2} + C

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Previous page : We have only one function, but one we can differentiate
Next page : The integral of sin(x)cos(x)dxLast modified: May 8, 2024 @ 08:50