4-x-2-x-dx-

Question Number 136948 by leena12345 last updated on 28/Mar/21

\int \frac{\sqrt{4 + x^{2}}}{x} d x

Answered by Olaf last updated on 28/Mar/21

F (x) = \int \frac{\sqrt{4 + x^{2}}}{x} d x

F (u) \overset{x = 2 sh u}{=} \int \frac{\sqrt{4 + {4 sh}^{2} u}}{2 sh u} 2 ch u d u

F (u) = \int \frac{2 ch u}{2 sh u} 2 ch u d u

F (u) = 2 \int \frac{{ch}^{2} u}{sh u} d u

F (u) = 2 \int \frac{{sh}^{2} u + 1}{sh u} d u

F (u) = 2 \int (sh u + \frac{1}{sh u}) d u

F (u) = 2 \int (sh u - \frac{2 e^{u}}{1 - e^{2 u}}) d u

F (u) = 2 [ch u - 2 argth (e^{u})]

F (u) = 2 [\sqrt{{sh}^{2} u + 1} - 2 \times \frac{1}{2} \ln ∣ \frac{1 + e^{u}}{1 - e^{u}} ∣]

F (u) = 2 [\sqrt{{sh}^{2} u + 1} - 2 \times \frac{1}{2} \ln ∣ \frac{1 + shu + ch u}{1 - sh u - ch u} ∣]

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

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

Answered by Dwaipayan Shikari last updated on 28/Mar/21

x = 2 s i n h u \Rightarrow {x e}^{u} = e^{2 u} - 1 \Rightarrow e^{u} = \frac{x \pm \sqrt{x^{2} + 4}}{2}

= \int \frac{4 {c o s h}^{2} (u)}{s i n h (u)} d u = 4 \int \frac{1 + {s i n h}^{2} u}{s i n h u} d u

= 4 \int \frac{1}{e^{u} - e^{- u}} d u + 4 c o s h (u) = 2 l o g (\frac{e^{u} - 1}{e^{u} + 1}) + 4 c o s h (u) + C

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

Answered by mathmax by abdo last updated on 28/Mar/21

Φ = \int \frac{\sqrt{4 + x^{2}}}{x} dx we do the changement x = 2 sht \Rightarrow

Φ = \int \frac{2 ch (t)}{2 sh (t)} (2 cht) dt = 2 \int \frac{{ch}^{2} t}{sht} dt

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

=_{e^{t} = y} \int \frac{2 + y^{2} + y^{- 2}}{y - y^{- 1}} \frac{dy}{y} = \int \frac{2 + y^{2} + y^{- 2}}{y^{2} - 1} dy

= \int \frac{{2 y}^{2} + y^{4} + 1}{y^{4} - y^{2}} dy = \int \frac{y^{4} + {2 y}^{2} + 1}{y^{2} (y^{2} - 1)} dy let decompose

F (y) = \frac{y^{4} + {2 y}^{2} + 1}{y^{4} - y^{2}} = \frac{y^{4} - y^{2} + {3 y}^{2} + 1}{y^{4} - y^{2}} = 1 + \frac{{3 y}^{2} + 1}{y^{4} - y^{2}}

v (y) = \frac{{3 y}^{2} + 1}{y^{2} (y^{2} - 1)} = \frac{{3 y}^{2} + 1}{y^{2} (y - 1) (y + 1)} = \frac{a}{y} + \frac{b}{y^{2}} + \frac{c}{y - 1} + \frac{d}{y + 1}

b = - 1, c = \frac{4}{2} = 2, d = \frac{4}{- 2} = - 2

v (y) = \frac{a}{y} - \frac{1}{y^{2}} + \frac{2}{y - 1} - \frac{2}{y + 1}

v (2) = \frac{13}{8} = \frac{a}{2} - \frac{1}{4} + 2 - \frac{2}{3} = \frac{a}{2} - \frac{1}{4} + \frac{4}{3} = \frac{a}{2} + \frac{13}{12} \Rightarrow a = \dots

\int F (y) dy = y + aln ∣ y ∣ + \frac{1}{y} + 2 \ln ∣ \frac{y - 1}{y + 1} ∣ + C

y = e^{t} ant t = argsh (\frac{x}{2}) = \ln (\frac{x}{2} + \sqrt{1 + \frac{x^{2}}{4}}) \Rightarrow

\int F (y) dy = \frac{x}{2} + \sqrt{1 + \frac{x^{2}}{4}} + aln (\frac{x}{2} + \sqrt{1 + \frac{x^{2}}{2}})

+ 2 \ln ∣ \frac{\frac{x}{2} - 1 + \sqrt{1 + \frac{x^{2}}{4}}}{\frac{x}{2} + 1 + \sqrt{1 + \frac{x^{2}}{4}}} ∣ + C = Φ