Question 31 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 22: x (log⁡\log x)^2

Correct Answer

x22(log⁡x)2−x22log⁡x+x24+C\frac{x^2}{2} (\log x)^2 - \frac{x^2}{2} \log x + \frac{x^2}{4} + C

Exercise: EXERCISE 7.6 | Q: 14 | (Page 39)
For More Understanding

Explanation

This question requires integration by parts. Following the ILATE rule, (log⁡x)2(\log x)^2 (logarithmic) is taken as the first function and xx (algebraic) as the second function. The integration is performed in two stages, similar to the examples shown in the textbook context for integration by parts with logarithmic functions.

Solution Steps

  1. Step 1: Let I=∫x(log⁡x)2dxI = \int x (\log x)^2 dx. Using integration by parts with first function u=(log⁡x)2u = (\log x)^2 and second function v=xv = x:

  2. Step 2: I=(log⁡x)2⋅x22−∫2log⁡xx⋅x22dx=x2(log⁡x)22−∫xlog⁡x dxI = (\log x)^2 \cdot \frac{x^2}{2} - \int \frac{2 \log x}{x} \cdot \frac{x^2}{2} dx = \frac{x^2 (\log x)^2}{2} - \int x \log x \, dx

  3. Step 3: For ∫xlog⁡x dx\int x \log x \, dx, apply integration by parts again with first function log⁡x\log x and second function xx:

  4. Step 4: ∫xlog⁡x dx=log⁡x⋅x22−∫1x⋅x22dx=x2log⁡x2−x24\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2 \log x}{2} - \frac{x^2}{4}

  5. Step 5: Substituting back: I=x2(log⁡x)22−x2log⁡x2+x24+CI = \frac{x^2 (\log x)^2}{2} - \frac{x^2 \log x}{2} + \frac{x^2}{4} + C

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