Question:
\[ \int\left(\frac{x-3}{x^2+9}\right)^2 d x= \]
\[ \int\left(\frac{x-3}{x^2+9}\right)^2 d x= \]
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Expand the numerator, split into two integrals and integrate each.
Updated On: Nov 13, 2026
- \(\frac{1}{3}\mathrm{tan}^{-1}(\frac{x}{3})-\frac{3}{x^{2}+9}+\mathrm{c}\), where c is the constant of integration.
- \(\frac{1}{3}\mathrm{tan}^{-1}(\frac{x}{3})-\frac{1}{x^{2}+9}+c\), where \(c\) is the constant of integration.
- \(\frac{1}{3}\mathrm{tan}^{-1}(\frac{x}{3})+\frac{3}{x^{2}+9}+c\), where \(c\) is the constant of integration.
- \(\frac{1}{3}\mathrm{tan}^{-1}(\frac{x}{3})+\frac{1}{x^{2}+9}+c\), where \(c\) is the constant of integration.
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The Correct Option is C
Solution and Explanation
Step 1: Expand
\((x-3)^2 = x^2 - 6x + 9 = (x^2 + 9) - 6x\).
Step 2: Split
\(\displaystyle\int\left[\dfrac{1}{x^2 + 9} - \dfrac{6x}{(x^2 + 9)^2}\right]dx\).
Step 3: Integrate each part
\(\displaystyle\int\dfrac{dx}{x^2 + 9} = \dfrac{1}{3}\tan^{-1}\dfrac{x}{3}\).
For \(\displaystyle\int\dfrac{6x}{(x^2+9)^2}dx\) put \(u = x^2 + 9\): \(\int\dfrac{3\,du}{u^2} = -\dfrac{3}{x^2 + 9}\).
Step 4: Combine
Total \(= \dfrac{1}{3}\tan^{-1}\dfrac{x}{3} + \dfrac{3}{x^2 + 9} + c\).
Final Answer:
The integral equals the inverse tangent term plus 3 over (x squared plus 9). \[ \boxed{\text{(C) }\dfrac{1}{3}\tan^{-1}\left(\dfrac{x}{3}\right) + \dfrac{3}{x^{2}+9} + c} \]
\((x-3)^2 = x^2 - 6x + 9 = (x^2 + 9) - 6x\).
Step 2: Split
\(\displaystyle\int\left[\dfrac{1}{x^2 + 9} - \dfrac{6x}{(x^2 + 9)^2}\right]dx\).
Step 3: Integrate each part
\(\displaystyle\int\dfrac{dx}{x^2 + 9} = \dfrac{1}{3}\tan^{-1}\dfrac{x}{3}\).
For \(\displaystyle\int\dfrac{6x}{(x^2+9)^2}dx\) put \(u = x^2 + 9\): \(\int\dfrac{3\,du}{u^2} = -\dfrac{3}{x^2 + 9}\).
Step 4: Combine
Total \(= \dfrac{1}{3}\tan^{-1}\dfrac{x}{3} + \dfrac{3}{x^2 + 9} + c\).
Final Answer:
The integral equals the inverse tangent term plus 3 over (x squared plus 9). \[ \boxed{\text{(C) }\dfrac{1}{3}\tan^{-1}\left(\dfrac{x}{3}\right) + \dfrac{3}{x^{2}+9} + c} \]
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