Question 28 of 33intermediate🔧 ApplyShort Answer2 marks

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

Correct Answer

x22log⁡\frac{x^2}{2} \log 2x - x24\frac{x^2}{4} + C

Exercise: EXERCISE 7.6 | Q: 5 | (Page 39)
For More Understanding

Explanation

The solution applies the integration by parts method, which is demonstrated in Example 19 and the text following Example 38 in the provided context. Following the ILATE rule, the logarithmic function log⁡2x\log 2x is chosen as the first function and the algebraic function xx as the second function. The derivative of log⁡2x\log 2x is found using the chain rule, consistent with the differentiation steps shown in the context.

Solution Steps

  1. Step 1: Let I=∫xlog⁡2x dxI = \int x \log 2x \, dx. Using integration by parts, take the first function as log⁡2x\log 2x and the second function as xx.

  2. Step 2: Apply the formula ∫u⋅v dx=u∫v dx−∫(dudx∫v dx)dx\int u \cdot v \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx.

  3. Step 3: Substitute the values: ∫x dx=x22\int x \, dx = \frac{x^2}{2} and ddx(log⁡2x)=1x\frac{d}{dx}(\log 2x) = \frac{1}{x}. This gives I=(log⁡2x)x22−∫1x⋅x22 dxI = (\log 2x) \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} \, dx.

  4. Step 4: Simplify the integral: I=x22log⁡2x−12∫x dxI = \frac{x^2}{2} \log 2x - \frac{1}{2} \int x \, dx.

  5. Step 5: Integrate the remaining term: I=x22log⁡2x−12⋅x22+C=x22log⁡2x−x24+CI = \frac{x^2}{2} \log 2x - \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{2} \log 2x - \frac{x^2}{4} + C.

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