Question 32 of 33intermediate🔧 ApplyShort Answer2 marks

Integrate the functions in Exercises 1 to 22: (x2x^{2}+1) log⁡\log x

Correct Answer
∫(x2+1)log⁡x dx=(x33+x)log⁡x−x39−x+C\int (x^2+1) \log x \, dx = \left(\frac{x^3}{3} + x\right) \log x - \frac{x^3}{9} - x + C
Exercise: EXERCISE 7.6 | Q: 15 | (Page 39)
For More Understanding

Explanation

This question requires integration by parts. The context shows the integration by parts formula applied in examples like ∫(log⁡x⋅1)dx\int (\log x \cdot 1) dx and ∫xexdx\int x e^x dx. For this problem, we take log⁡x\log x as the first function and (x2+1)(x^2+1) as the second function, following the ILATE rule where logarithmic functions come first.

Solution Steps

  1. Step 1: Apply integration by parts taking log⁡x\log x as first function and (x2+1)(x^2+1) as second function.

  2. Step 2: Using ∫u⋅v dx=u∫v dx−∫[ddx(u)∫v dx]dx\int u \cdot v \, dx = u \int v \, dx - \int \left[\frac{d}{dx}(u) \int v \, dx\right] dx, we get:

  3. Step 3: ∫(x2+1)log⁡x dx=log⁡x∫(x2+1)dx−∫[1x∫(x2+1)dx]dx\int (x^2+1) \log x \, dx = \log x \int (x^2+1) dx - \int \left[\frac{1}{x} \int (x^2+1) dx\right] dx

  4. Step 4: Calculate ∫(x2+1)dx=x33+x\int (x^2+1) dx = \frac{x^3}{3} + x

  5. Step 5: Substitute: =(x33+x)log⁡x−∫1x(x33+x)dx= \left(\frac{x^3}{3} + x\right) \log x - \int \frac{1}{x} \left(\frac{x^3}{3} + x\right) dx

  6. Step 6: Simplify: =(x33+x)log⁡x−∫(x23+1)dx= \left(\frac{x^3}{3} + x\right) \log x - \int \left(\frac{x^2}{3} + 1\right) dx

  7. Step 7: Final answer: =(x33+x)log⁡x−x39−x+C= \left(\frac{x^3}{3} + x\right) \log x - \frac{x^3}{9} - x + C

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