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Evaluate\(\int\limits_{0}^{1}\cfrac{2\text x}{1+\text x^2}d\text x \) |
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Answer» Let I = \(\int\limits_{0}^{1}\cfrac{2\text x}{1+\text x^2}d\text x \) Substituting 1+x2 = t ⇒ 2x dx=dt Also, When x = 0, t=1 and x=1, t = 2 We get I = \(\int\limits_{0}^{1}\cfrac{1}tdt\) = loget \(|^2_1\) = loge 2 - loge1 = loge2 (Since loge1 = 0) |
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