x-2-4x-5-dx-

Question Number 123234 by benjo_mathlover last updated on 24/Nov/20

\int \sqrt{x^{2} - 4 x + 5} d x

Commented by liberty last updated on 24/Nov/20

Answered by MJS_new last updated on 24/Nov/20

\int \sqrt{x^{2} - 4 x + 5} d x =

[t = x - 2 + \sqrt{x^{2} - 4 x + 5} \to d x = \frac{\sqrt{x^{2} - 4 x + 5}}{x - 2 + \sqrt{x^{2} - 4 x + 5}}]

[use x = \frac{t^{2} + 4 t - 1}{2 t} \land t > 0]

= \int \frac{{(t^{2} + 1)}^{2}}{4 t^{3}} d t = \int (\frac{t}{4} + \frac{1}{2 t} + \frac{1}{4 t^{3}}) d t =

= \frac{t^{4} - 1}{8 t^{2}} + \frac{1}{2} \ln t =

= \frac{x - 2}{2} \sqrt{x^{2} - 4 x + 5} + \frac{1}{2} \ln (x - 2 + \sqrt{x^{2} - 4 x + 5}) + C

Answered by Dwaipayan Shikari last updated on 24/Nov/20

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

= i \int \sqrt{1 - t^{2}} d t = i \int c o s θ \sqrt{1 - {s i n}^{2} θ} d θ

= i (\frac{θ}{2} + \frac{s i n 2 θ}{4}) = i (\frac{{s i n}^{- 1} t}{2} + \frac{t \sqrt{1 - t^{2}}}{2}) = i (\frac{{s i n}^{- 1} (\frac{x - 2}{i})}{2} + \frac{\frac{(x - 2)}{i} \sqrt{1 + {(x - 2)}^{2}}}{2})

= i \frac{{s i n}^{- 1} (\frac{x - 2}{i})}{2} + \frac{(x - 2) \sqrt{x^{2} - 4 x + 5}}{2}

= \frac{1}{2} l o g (x - 2 + \sqrt{x^{2} - 4 x + 5}) + \frac{(x - 2)}{2} \sqrt{x^{2} - 4 x + 5}

Commented by Dwaipayan Shikari last updated on 24/Nov/20

{i s i n}^{- 1} \frac{(x - 2)}{i} = t

\frac{(x - 2)}{i} = s i n (\frac{t}{i}) \Rightarrow (\frac{x - 2}{i}) = \frac{e^{t} - e^{- t}}{2 i} \Rightarrow (2 x - 4) = e^{t} - e^{- t}

e^{t} - e^{- t} = (2 x - 4)

a - \frac{1}{a} = (2 x - 4) \Rightarrow a^{2} - (2 x - 4) a - 1 = 0 \Rightarrow a = \frac{(2 x - 4) + \sqrt{{(2 x - 4)}^{2} + 4}}{2}

a = x - 2 + \sqrt{x^{2} - 4 x + 5} = e^{t}

t = l o g (x - 2 + \sqrt{x^{2} - 4 x + 5})

Answered by mathmax by abdo last updated on 24/Nov/20

A = \int \sqrt{x^{2} - 4 x + 5} dx

x^{2} - 4 x + 5 = x^{2} - 4 x + 4 + 1 = {(x - 2)}^{2} + 1 we do the changement

x - 2 = sht \Rightarrow A = \int \sqrt{{(x - 2)}^{2} + 1} dx = \int \sqrt{1 + {sh}^{2}} t ch (t) dt

= \int {ch}^{2} t dt = \int \frac{ch (2 t) + 1}{2} dt = \frac{t}{2} + \frac{1}{4} sh (2 t) + C

= \frac{t}{2} + \frac{1}{2} sh (t) ch (t) + C = \frac{1}{2} argsh (x - 2) + \frac{x - 2}{2} \sqrt{1 + {(x - 2)}^{2}} + C

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