dx-x-3-x-2-

Question Number 121601 by benjo_mathlover last updated on 09/Nov/20

\int \frac{dx}{x \sqrt{3 + x^{2}}} ?

Answered by liberty last updated on 10/Nov/20

let x = \sqrt{3} \tan γ

I = \int \frac{\sqrt{3} \sec^{2} γ}{\sqrt{3} \tan γ \sqrt{3 + 3 \tan^{2} γ}} d γ

I = \frac{1}{\sqrt{3}} \int \frac{\sec γ}{\tan γ} d γ = \frac{1}{\sqrt{3}} \int \frac{1}{\sin γ} d γ

I = \frac{1}{\sqrt{3}} \int cosec γ d γ = \frac{1}{\sqrt{3}} \ln ∣ cosec γ - \cot γ ∣ + c

I = \frac{1}{\sqrt{3}} \ln ∣ \frac{\sqrt{3 + x^{2}} - \sqrt{3}}{x} ∣ + c

Answered by Olaf last updated on 10/Nov/20

x = \sqrt{3} \sinh u

I = \int \frac{\sqrt{3} \cosh u}{\sqrt{3} \sinh u \sqrt{3 + {3 \sinh}^{2} u}} d u

I = \frac{1}{\sqrt{3}} \int \frac{d u}{\sinh u}

I = \frac{1}{\sqrt{3}} \int \frac{d u}{\frac{e^{u} - e^{- u}}{2}}

I = \frac{2}{\sqrt{3}} \int \frac{e^{u} d u}{e^{2 u} - 1}

I = \frac{1}{\sqrt{3}} \int [\frac{1}{e^{u} - 1} - \frac{1}{e^{u} + 1}] e^{u} d u

I = \frac{1}{\sqrt{3}} \ln ∣ \frac{e^{u} - 1}{e^{u} + 1} ∣

I = \frac{1}{\sqrt{3}} \ln ∣ \frac{e^{u / 2} - e^{- u / 2}}{e^{u / 2} + e^{- u / 2}} ∣

I = \frac{1}{\sqrt{3}} \ln ∣ \tanh (\frac{u}{2}) ∣

I = \frac{1}{\sqrt{3}} \ln ∣ \tanh (\frac{1}{2} argsinh (\frac{x}{\sqrt{3}})) ∣

Answered by Dwaipayan Shikari last updated on 10/Nov/20

\int \frac{d x}{x \sqrt{3 + x^{2}}}

\int \frac{{s e c}^{2} θ}{\sqrt{3} t a n θ s e c θ} d θ x = \sqrt{3} t a n θ

= \frac{1}{\sqrt{3}} \int \frac{1}{s i n θ} d θ = \frac{1}{3} l o g (t a n \frac{θ}{2}) + C

= \frac{1}{\sqrt{3}} l o g (\frac{- \sqrt{3} + \sqrt{3 + x^{2}}}{x}) + C

Answered by TANMAY PANACEA last updated on 10/Nov/20

\int \frac{x d x}{x^{2} \sqrt{3 + x^{2}}}

t^{2} = 3 + x^{2} \to 2 t d t = x d x

\int \frac{t d t}{(t^{2} - 3) \times t} = \int \frac{d t}{t^{2} - 3}

= \frac{1}{2 \sqrt{3}} l n (\frac{t - \sqrt{3}}{t + \sqrt{3}}) + c

= \frac{1}{2 \sqrt{3}} l n (\frac{\sqrt{3 + x^{2}} - \sqrt{3}}{\sqrt{3 + x^{2}} + \sqrt{3}}) + c