MHT CET · Maths · Indefinite Integration
\(\int \frac{d x}{x^{2}+4 x+13}=\)
- A \(\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)+c\)
- B \(\frac{1}{6} \log \left(\frac{x-1}{x+5}\right)+c\)
- C \(\frac{1}{6} \tan ^{-1}\left(\frac{x+2}{3}\right)+c\)
- D \(3 \tan ^{-1}\left(\frac{x+2}{3}\right)+c\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3} \tan ^{-1}\left(\frac{x+2}{3}\right)+c\)
Step-by-step Solution
Detailed explanation
(D)
\(\int \frac{1}{x^{2}+4 x+13} d x\)
\(=\int \frac{1}{x^{2}+4 x+4+9} d x=\int \frac{1}{(x+2)^{2}+3^{2}} d x=\frac{1}{3} \tan ^{-1}\left(\frac{x-2}{3}\right)-c\)
\(\int \frac{1}{x^{2}+4 x+13} d x\)
\(=\int \frac{1}{x^{2}+4 x+4+9} d x=\int \frac{1}{(x+2)^{2}+3^{2}} d x=\frac{1}{3} \tan ^{-1}\left(\frac{x-2}{3}\right)-c\)
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