dx-x-x-2-2x-2-

Question Number 157442 by bobhans last updated on 23/Oct/21

\int \frac{dx}{x - \sqrt{x^{2} + 2 x + 2}}

Answered by MJS_new last updated on 23/Oct/21

\int \frac{d x}{x - \sqrt{x^{2} + 2 x + 2}} =

[t = x + 1 + \sqrt{x^{2} + 2 x + 2} \to d x = \frac{\sqrt{x^{2} + 2 x + 2}}{x + 1 + \sqrt{x^{2} + 2 x + 2}} d t]

= - \frac{1}{2} \int \frac{t^{2} + 1}{t (t + 1)} d t = \int (\frac{1}{t + 1} - \frac{1}{2 t} - \frac{1}{2}) d t =

= \ln (t + 1) - \frac{1}{2} \ln t - \frac{1}{2} t =

= \frac{1}{2} (\ln \frac{{(t + 1)}^{2}}{t} - t) =

= \frac{1}{2} (\ln (1 + \sqrt{x^{2} + 2 x + 2}) - x - \sqrt{x^{2} + 2 x + 2}) + C

Answered by cortano last updated on 23/Oct/21

\frac{1}{x - \sqrt{x^{2} + 2 x + 2}} = \frac{x + \sqrt{x^{2} + 2 x + 2}}{- 2 x - 2}

I = - \int \frac{x}{2 x + 2} d x - \int \frac{\sqrt{x^{2} + 2 x + 2}}{2 x + 2} d x

I_{1} = - \frac{1}{2} \int \frac{x + 1 - 1}{x + 1} d x = - \frac{1}{2} x + \ln ∣ x + 1 ∣ + c_{1}

I_{2} = - \frac{1}{2} \int \frac{\sqrt{{(x + 1)}^{2} + 1}}{x + 1} d x

x + 1 = \tan u

I_{2} = - \frac{1}{2} \int \frac{\sec u}{\tan u} \sec^{2} d u

I_{2} = - \frac{1}{2} \int c s c x \sec^{2} x d x

I B P \Rightarrow {\begin{cases} u = c s c x \Rightarrow d u = - c s c x \cot x d x \\ v = \tan x \end{cases}

I_{2} = - \frac{1}{2} {c s c x \tan x + \int c s c x d x}

I_{2} = - \frac{1}{2} c s c x \tan x - \frac{1}{2} \ln ∣ c s c x - \cot x ∣ + c_{2}

I = I_{1} + I_{2}

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