Question

∫[1/{1 + (logx)²}]d(logx)dx = _______ + c 

(a) [{tan^–1(log x)}/x] 

(b) tan^–1(logx) 

(c) [(tan^–1x)/x] 

(d) tan^–1x

Answer 1

 The correct option (b) tan^–1(log x)

Explanation:

I = ∫[{d(logx)dx}/{1 + (logx)²}] 

Let logx = t 

d(logx)dx = dt 

∴ I = ∫[dt/(1 + t²)] 

= tan^–1t + c 

= tan^–1(logx) + c