Use integration to find $$\int \frac{1}{x^3} \ln x \, dx$$ (b) Hence calculate $$\int_1^2 \frac{1}{x^3} \ln x \, dx$$ - Edexcel - A-Level Maths Pure - Question 14 - 2013 - Paper 1

Now, we can use the previous result to evaluate the definite integral:

$\int_1^2 \frac{1}{x^3} \ln x \, dx = \left[ -\frac{1}{2x^2} \ln x + \frac{1}{4x^2} \right]_1^2$

Calculating the limit at 2:

$= -\frac{1}{2(2^2)} \ln(2) + \frac{1}{4(2^2)}$ $= -\frac{1}{8} \ln(2) + \frac{1}{16}$

Calculating the limit at 1:

$= -\frac{1}{2(1^2)} \ln(1) + \frac{1}{4(1^2)}$ $= 0 + \frac{1}{4}$

Subtracting these results:

$= \left(-\frac{1}{8} \ln(2) + \frac{1}{16}\right) - \left(0 + \frac{1}{4}\right)$

Resulting in:

$= -\frac{1}{8} \ln(2) + \frac{1}{16} - \frac{4}{16}$ $= -\frac{1}{8} \ln(2) - \frac{3}{16}$

This gives us the final computed value for the definite integral.