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