MHT CET · Maths · Definite Integration
\(\int_{0}^{1}\left(\frac{x^{2}-2}{x^{2}+1}\right) d x=\)
- A \(1+\frac{3 \pi}{4}\)
- B \(1-\frac{3 \pi}{4}\)
- C \(1-\frac{\pi}{4}\)
- D \(1+\frac{\pi}{4}\)
Answer & Solution
Correct Answer
(B) \(1-\frac{3 \pi}{4}\)
Step-by-step Solution
Detailed explanation
(C)
\(\int_{0}^{1} \frac{x^{2}+1-3}{x^{2}+1} d x =\int_{0}^{1} \frac{x^{2}+1}{x^{2}+1}-\frac{3}{x^{2}+1} d x \)
\( =\int_{0}^{1} 1-\frac{3}{x^{2}+1} d x=\left[x-3 \tan ^{-1} x\right]_{0}^{1} \)
\( =\left(1-3 \tan ^{-1} 1\right)-\left(0-3 \tan ^{-1} 0\right)=1-3 \frac{\pi}{4}\)
\(\int_{0}^{1} \frac{x^{2}+1-3}{x^{2}+1} d x =\int_{0}^{1} \frac{x^{2}+1}{x^{2}+1}-\frac{3}{x^{2}+1} d x \)
\( =\int_{0}^{1} 1-\frac{3}{x^{2}+1} d x=\left[x-3 \tan ^{-1} x\right]_{0}^{1} \)
\( =\left(1-3 \tan ^{-1} 1\right)-\left(0-3 \tan ^{-1} 0\right)=1-3 \frac{\pi}{4}\)
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