Question

$\int\left(\log _{e}(1+\tan x)+\frac{2 x}{1+\cos 2 x+\sin 2 x}\right) d x$ is equal to

(A) $x \ell n (1+\tan x)+c$

(B) $x^{2} \ell n (1+\cos x)$

(C) $x \ell n (1+\sec x)$

(D) None

Answer 1

$\begin{aligned}& \int \log (1+\tan x)+\int \frac{2 x d x}{1+\cos 2 x+\sin 2 x} \\=& \log (1+\tan x) \int 1 \cdot d x-\int x \frac{\sec ^{2} x}{1+\tan x} d x+\int \frac{2 x d x}{1+\cos 2 x+\sin 2 x} \\=& x \ln (1+\tan x)-\int \frac{2 x}{1+\cos 2 x+\sin 2 x} d x+\int \frac{2 x d x}{1+\cos 2 x+\sin 2 x} \\=& x \ln (1+\tan x)+c\end{aligned}$

So, The correct answer of this question will be option (A).