The correct option is A π/4x+x^ 1x 1/2log(1+x^2)+c cosecθ =√(1+x^2) sin θ =1√(1+x^2) ∫ cot^ 1(x 1x+1)dx x=cotθ ∫ cot^ 1(cotθ 1co ...

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Complex Numbers

∫ cot-1x-1/x+...

Question

$\int c o t^{- 1} (\frac{x - 1}{x + 1}) d x =$

\frac{π}{4} x + x {cot}^{- 1} x - \frac{1}{2} log (1 + x^{2}) + c

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x {cot}^{- 1} x + \frac{1}{2} log (1 + x^{2}) + c

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\frac{π}{4} x - \frac{1}{2} log (1 + x^{2}) + c

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^{\frac{π}{4} x + x {cot}^{- 1} x} + \frac{1}{2} log (1 + x^{2}) + c

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Solution

The correct option is A $\frac{π}{4} x + x {cot}^{- 1} x - \frac{1}{2} log (1 + x^{2}) + c$
$c o s e c θ = \sqrt{1 + x^{2}}$
$s i n θ = \frac{1}{\sqrt{1 + x^{2}}}$
$\int c o t^{- 1} (\frac{x - 1}{x + 1}) d x$
$x = c o t θ$
$\int c o t^{- 1} (\frac{c o t θ - 1}{c o t θ + 1}) [c o s e c^{2} θ] d θ$
$- \int c o t^{- 1} [c o t (π / 4 + θ)] (c o s e c^{2} θ) d θ$
$= - \int (π / 4 + θ (c o s e c^{2} θ)) d θ$
$= - [\int π / 4 c o s e c^{γ} θ d θ + \int θ c o s e c^{γ θ d θ}]$
$= - [- π / 4 c o t θ + [θ (- c o t θ) + \int c o t θ d θ]$
$= π / 4 c o t θ + c o t θ - l o g | s i n θ | + c$
$= π / 4 x + x c o t^{- 1} x - 1 / 2 l o g (\sqrt{1 + x^{2}}) + c$
$\int c o t^{- 1} (\frac{x - 1}{x + 1}) d x = π / 4 x + x c o t^{- 1} x - 1 / 2 l o g (\sqrt{1 + x^{2}}) + c$

Suggest Corrections

Similar questions

\int \frac{x^{2} {tan}^{- 1} x}{1 + x^{2}} d x = {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2}) + f (x) + c

f (x) =

\int \frac{x d x}{(1 + x) (1 + x^{2})} = - \frac{1}{2} log (1 + x) + \frac{1}{4} log (1 + x^{2}) + \frac{1}{2} {tan}^{- 1} x

\int \frac{d x}{(1 + x) (1 + x^{2})} = \frac{1}{2} log (1 + x) + \frac{1}{4} log (1 + x^{2}) + \frac{1}{2} {tan}^{- 1} x

({sin}^{- 1} x + \frac{1}{2} log \frac{1 + x}{1 - x})

{sin}^{- 1} \frac{2 x}{1 + x^{2}}; {cos}^{- 1} \frac{1 - x^{2}}{1 + x^{2}}; {tan}^{- 1} \frac{2 x}{1 - x^{2}} .

2 {tan}^{- 1} x .

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

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