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Exercise 11.9 · Q2

Q.ex(x−12x2)e^{x}\left(\dfrac{x-1}{2x^{2}}\right)

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✓ Free question

Splitting the single fraction into two pieces reveals one term is a function and the other its own derivative.

Step 1. Split the fraction. x−12x2=x2x2−12x2=12x−12x2\dfrac{x-1}{2x^2}=\dfrac{x}{2x^2}-\dfrac1{2x^2}=\dfrac1{2x}-\dfrac1{2x^2}.

Step 2. Identify f(x)f(x). Take f(x)=12xf(x)=\dfrac1{2x}; then f′(x)=−12x2f'(x)=-\dfrac1{2x^2}, matching the second term exactly.

Step 3. Match the pattern and apply Result 11.2. ex(12x−12x2)=ex[f(x)+f′(x)]e^x\left(\dfrac1{2x}-\dfrac1{2x^2}\right)=e^x[f(x)+f'(x)], so ∫ex(x−12x2)dx=exf(x)+c=ex2x+c\displaystyle\int e^x\left(\dfrac{x-1}{2x^2}\right)dx=e^xf(x)+c=\dfrac{e^x}{2x}+c.

Step 4. Check. ddx[ex2x]=ex2x−ex2x2=ex(12x−12x2)=ex⋅x−12x2\dfrac d{dx}\left[\dfrac{e^x}{2x}\right]=\dfrac{e^x}{2x}-\dfrac{e^x}{2x^2}=e^x\left(\dfrac1{2x}-\dfrac1{2x^2}\right)=e^x\cdot\dfrac{x-1}{2x^2}, matching the integrand.

✓Final answer

ex2x+c\dfrac{e^x}{2x}+c

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