1.

Evaluate\(\int\limits_{0}^{1}\cfrac{2\text x}{1+\text x^2}d\text x \)

Answer»

Let I =  \(\int\limits_{0}^{1}\cfrac{2\text x}{1+\text x^2}d\text x \)

Substituting 1+x= 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)



Discussion

No Comment Found