∫ 1 + tan 2 x 1 - tan 2 x d x =

1+tan2x1-tan2xdx=
  1. log1-tanx1+tanx+c
  2. log1+tanx1-tanx+c
  3. 12log1-tanx1+tanx+c
  4. 12log1+tanx1-tanx+c

Solution

Let,

I=1+tan2x1-tan2xdx

I=sec2x1-tan2xdx

Put tanx=t, we get

sec2x dx=dt

I=dt1-t2

I=12log 1+t1-t+c

I=12log 1+tanx1-tanx+c.

Asked in: AP EAMCET 2020 (23 Sep Shift 1)

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