Evaluate -1x2-2x-2 dx

The correct option is A tan^ 1(x+1)+C ∫1x^2+2x+2dx = ∫1(x^2+1+2x)+1 = ∫1(x+1)^2+1dx Let 1+x= μ ,du=dx ∫11+μ^2du #160; = tan^ 1u ...

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Theorems on Integration

Evaluate ∫1...

Question

Evaluate $\int \frac{1}{x^{2} + 2 x + 2} d x$

{tan}^{- 1} (x + 1) + C

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{tan}^{- 1} (x + 2) + C

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{tan}^{- 1} (x + 3) + C

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None of these

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

\int \frac{x^{2} {tan}^{- 1} x}{1 + x^{2}}

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

\frac{3 π}{4}

\frac{8}{31}

(\frac{1 - x}{1 + x}) - \frac{1}{2}

\frac{π}{12}

(\frac{8}{79})

\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}

\tan^{- 1} (\frac{x - 2}{x - 4}) + \tan^{- 1} (\frac{x + 2}{x + 4}) = \frac{π}{4}

\tan^{- 1} (2 + x) + \tan^{- 1} (2 - x) = \tan^{- 1} \frac{2}{3}, where x < - \sqrt{3} or, x > \sqrt{3}

\tan^{- 1} \frac{x - 2}{x - 1} + \tan^{- 1} \frac{x + 2}{x + 1} = \frac{π}{4}

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