MHT CET · Maths · Indefinite Integration
\(\int\left(\frac{x-3}{x^2+9}\right)^2 \mathrm{~d} x=\)
- A \(\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{3}{x^2+9}+c\), where \(c\) is the constant of integration.
- B \(\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{1}{x^2+9}+c\), where \(c\) is the constant of integration.
- C \(\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+\frac{3}{x^2+9}+c\), where \(c\) is the constant of integration.
- D \(\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)-\frac{1}{x^2+9}+c\), where \(c\) is the constant of integration.
Answer & Solution
Correct Answer
(C) \(\frac{1}{3} \tan ^{-1}\left(\frac{x}{3}\right)+\frac{3}{x^2+9}+c\), where \(c\) is the constant of integration.
Step-by-step Solution
Detailed explanation
\( \int\left(\frac{x-3}{x^2+9}\right)^2 \mathrm{~d} x=\int \frac{x^2-6x+9}{(x^2+9)^2} \mathrm{~d} x \) \( =\int \left( \frac{x^2+9}{(x^2+9)^2} - \frac{6x}{(x^2+9)^2} \right) \mathrm{~d} x \)
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