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Exercise 11.5 · Q1

Q.x3+4x2−3x+2x2\dfrac{x^3+4x^2-3x+2}{x^2}

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✓ Free question

Splitting the numerator over x2x^2 turns this into four separate power-of-xx integrals.

Step 1. Split the fraction. x3+4x2−3x+2x2=x+4−3x+2x2\dfrac{x^3+4x^2-3x+2}{x^2}=x+4-\dfrac{3}{x}+\dfrac{2}{x^2}.

Step 2. Integrate each piece. ∫x dx=x22\displaystyle\int x\,dx=\dfrac{x^2}{2}, ∫4 dx=4x\displaystyle\int 4\,dx=4x, ∫3xdx=3log⁡∣x∣\displaystyle\int\dfrac{3}{x}dx=3\log|x|, ∫2x2dx=−2x\displaystyle\int\dfrac{2}{x^2}dx=-\dfrac2x.

Step 3. Combine. ∫x3+4x2−3x+2x2dx=x22+4x−3log⁡∣x∣−2x+c\displaystyle\int\dfrac{x^3+4x^2-3x+2}{x^2}dx=\dfrac{x^2}{2}+4x-3\log|x|-\dfrac{2}{x}+c.

Step 4. Check. Differentiating back: ddx[x22+4x−3log⁡∣x∣−2x]=x+4−3x+2x2\dfrac{d}{dx}\left[\dfrac{x^2}2+4x-3\log|x|-\dfrac2x\right]=x+4-\dfrac3x+\dfrac2{x^2}, matching the split integrand of Step 1.

✓Final answer

x22+4x−3log⁡∣x∣−2x+c\dfrac{x^2}{2}+4x-3\log|x|-\dfrac{2}{x}+c

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