MHT CET · Maths · Indefinite Integration
\(\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x\) is equal to
- A \(\log \frac{27}{32}\)
- B \(\log \frac{32}{27}\)
- C \(\log \frac{8}{9}\)
- D \(\log \frac{3}{4}\)
Answer & Solution
Correct Answer
(B) \(\log \frac{32}{27}\)
Step-by-step Solution
Detailed explanation
Let \(I=\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x\)
\(\quad=\int_{5}^{10}\left[\frac{-1}{x-1}+\frac{1}{x-2}\right] d x\)
\(=[-\log (x-1)+\log (x-2)]_{5}^{10}\)
\(=-\log 9+\log 8+\log 4-\log 3\)
\(=-2 \log 3+3 \log 2+2 \log 2-\log 3\)
\(=-3 \log 3+5 \log 2\)
\(=-\log 27+\log 32\)
\(=\log \frac{32}{27}\)
\(\quad=\int_{5}^{10}\left[\frac{-1}{x-1}+\frac{1}{x-2}\right] d x\)
\(=[-\log (x-1)+\log (x-2)]_{5}^{10}\)
\(=-\log 9+\log 8+\log 4-\log 3\)
\(=-2 \log 3+3 \log 2+2 \log 2-\log 3\)
\(=-3 \log 3+5 \log 2\)
\(=-\log 27+\log 32\)
\(=\log \frac{32}{27}\)
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