indefinite-integration

Question: $\int{\frac{1+{{\tan }^{2}}x}{1-{{\tan }^{2}}x}\,dx}$ equals to

1) $\log \left( \frac{1-\tan x}{1+\tan x} \right)+c$

2) $\log \left( \frac{1+\tan x}{1-\tan x} \right)+c$

3) $\frac{1}{2}\log \left( \frac{1-\tan x}{1+\tan x} \right)+c$

4) $\frac{1}{2}\log \left( \frac{1+\tan x}{1-\tan x} \right)+c$

Solution:

Explanation: No Explanation

Integration by substitution

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