$\int {1 + 2 t a n x (t a n x + s e c x)}^{1 / 2} d x =$

l o g | s e c x (s e c x - t a n x) | + c

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l o g | s e c x (s e c x + t a n x) | + c

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l o g ∣ ∣ \frac{s e c x}{s e c x + t a n x} ∣ ∣ + c

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l o g | c o s x (s e c x + t a n x) | + c

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Solution

The correct option is B $l o g | s e c x (s e c x + t a n x) | + c$
$I = \int {1 + 2 t a n x (t a n x + s e c x)}^{1 / 2} d x = \int [1 + 2 t a n^{2} x + 2 t a n x s e c x]^{1 / 2} d x = \int (s e c^{2} x + t a n^{2} x + 2 s e c x t a n x)^{1 / 2} d x = \int (s e c x + t a n x) d x = l o g (s e c x + t a n x) + l o g s e c x + c = l o g [s e c x (s e c x + t a n x)] + c$

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Similar questions

\int {1 + 2 tan x (tan x + sec x)}^{\frac{1}{2}} d x

equals

A. x tan⁻¹ (x + 1) + C

B. tan^−
1 (x + 1) + C

C. (x + 1) tan⁻¹ x + C

D. tan⁻¹ x + C

\int {1 + 2 tan x (tan x + sec x)}^{\frac{1}{2}} d x =

I = \int \frac{(1 + {tan}^{2} x)^{2}}{1 + {tan}^{2} x} {sec}^{2} x d x =

\int \frac{\sec^{2} x}{1 - \tan^{2} x} d x

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