Match List I with List II LIST ILIST II A. ∫(sin x)/(1+cos x) dx I. e…

The correct answer is Option (2) → A-III, B-IV, C-I, D-II

(A) \(\int\frac{\sin x}{1+\cos x}\,dx\)

This matches (III) \(-\log|1+\cos x|+C\).

By substitution, put \(t = 1 + \cos x\), so \(dt = -\sin x\,dx\):

    \( \displaystyle \int\frac{\sin x}{1+\cos x}\,dx = -\int\frac{dt}{t} \)

    \( = -\log|1+\cos x| + C \)

(B) \(\int\frac{1}{1-\tan x}\,dx\)

This matches (IV) \(\frac{x}{2}-\frac{1}{2}\log|\cos x-\sin x|+C\).

Multiplying the numerator and the denominator by cos x:

    \( \displaystyle \frac{1}{1-\tan x} = \frac{\cos x}{\cos x - \sin x} \)

Write the numerator in terms of the denominator and \(\cos x + \sin x\), the negative of its derivative:

    \( \displaystyle \cos x = \frac{1}{2}\big[(\cos x - \sin x) + (\cos x + \sin x)\big] \)

So

    \( \displaystyle \int\frac{\cos x}{\cos x - \sin x}\,dx = \frac{1}{2}\int 1\,dx + \frac{1}{2}\int\frac{\cos x + \sin x}{\cos x - \sin x}\,dx \)

Since \(\frac{d}{dx}(\cos x - \sin x) = -(\cos x + \sin x)\), the second integral is \(-\log|\cos x - \sin x|\):

    \( \displaystyle = \frac{x}{2} - \frac{1}{2}\log|\cos x - \sin x| + C \)

(C) \(\int\frac{e^{\tan^{-1}x}}{1+x^2}\,dx\)

This matches (I) \(e^{\tan^{-1}x}+C\).

Put \(t = \tan^{-1}x\), so \(dt = \frac{dx}{1+x^2}\):

    \( \displaystyle \int\frac{e^{\tan^{-1}x}}{1+x^2}\,dx = \int e^{t}\,dt \)

    \( = e^{\tan^{-1}x} + C \)

(D) \(\int\frac{1}{x+x\log x}\,dx\)

This matches (II) \(\log(\log x+1)+C\).

The denominator is \(x(1 + \log x)\). Put \(t = 1 + \log x\), so \(dt = \frac{dx}{x}\):

    \( \displaystyle \int\frac{dx}{x(1+\log x)} = \int\frac{dt}{t} \)

    \( = \log(\log x + 1) + C \)