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What's the integral of ln(x)/x^2?

To evaluate the integral of ( \frac{\ln(x)}{x^2} ), we can utilize the method of integration by parts. The integral can be expressed as follows:

ln(x)x2dx\int \frac{\ln(x)}{x^2} \, dx

We choose our variables for integration by parts:

  • Let ( u = \ln(x) ), which implies ( du = \frac{1}{x} , dx ).
  • Let ( dv = \frac{1}{x^2} , dx ), which gives us ( v = -\frac{1}{x} ).

Applying the integration by parts formula, which is given by

udv=uvvdu,\int u \, dv = uv - \int v \, du,

we can substitute our choices into the formula:

ln(x)x2dx=ln(x)x1xdx.\int \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(x)}{x} - \int -\frac{1}{x} \, dx.

The integral ( \int -\frac{1}{x} , dx ) can be evaluated using the standard rule for the integral of ( \frac{1}{x} ):

1xdx=ln(x)+C.\int -\frac{1}{x} \, dx = -\ln(x) + C.

Now substituting this back into our previous expression, we have:

ln(x)x2dx=ln(x)x+ln(x)+C.\int \frac{\ln(x)}{x^2} \, dx = -\frac{\ln(x)}{x} + \ln(x) + C.

Next, we can simplify the expression:

ln(x)x2dx=1x+ln(x)x+C.\int \frac{\ln(x)}{x^2} \, dx = -\frac{1}{x} + \frac{\ln(x)}{x} + C.

Thus, the final result for the integral is:

ln(x)x2dx=1x+ln(x)x+C.\int \frac{\ln(x)}{x^2} \, dx = -\frac{1}{x} + \frac{\ln(x)}{x} + C.

In summary, the integral of ( \frac{\ln(x)}{x^2} ) is:

1x+ln(x)x+C.-\frac{1}{x} + \frac{\ln(x)}{x} + C.
Answered by: Prof. Alan Smith
A-Level Physics Tutor
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