find-dx-x-2-1-x-2-3-

Question Number 73482 by abdomathmax last updated on 13/Nov/19

f i n d \int \frac{d x}{\sqrt{x^{2} + 1} + \sqrt{x^{2} + 3}}

Commented by abdomathmax last updated on 17/Nov/19

l e t I = \int \frac{d x}{\sqrt{x^{2} + 1} + \sqrt{x^{2} + 3}} \Rightarrow

I = \int \frac{\sqrt{x^{2} + 3} - \sqrt{x^{2} + 1}}{2} d x = \frac{1}{2} \int \sqrt{x^{2} + 3} d x - \frac{1}{2} \int \sqrt{x^{2} + 1} d x

w e h a v e \int \sqrt{x^{2} + 3} d x =_{x = \sqrt{3} s h (t)} \sqrt{3} \int \sqrt{3} c h (t) c h (t)

= 3 \int \frac{1 + c h (2 t)}{2} d t = \frac{3}{2} t + \frac{3}{4} s h (2 t) + c_{1}

= \frac{3}{4} t + \frac{3}{4} 2 s h (t) c h (t) + c_{1}

= \frac{3}{4} a r g s h (\frac{x}{\sqrt{3}}) + \frac{3}{2} \frac{x}{\sqrt{3}} \sqrt{1 + \frac{x^{2}}{3}} + c

= \frac{3}{4} l n (\frac{x}{\sqrt{3}} + \sqrt{1 + \frac{x^{2}}{3}}) + \frac{x}{2} \sqrt{3 + x^{2}} + c_{1}

\int \sqrt{x^{2} + 1} d x =_{x = s h (t)} \int c h (t) c h (t) d t

= \int \frac{1 + c h (2 t)}{2} d t = \frac{t}{2} + \frac{1}{4} s h (2 t) + c_{2}

= \frac{t}{2} + \frac{1}{2} s h (g) c h (t) + c_{2}

= \frac{1}{2} l n (x + \sqrt{1 + x^{2}}) + \frac{x}{2} \sqrt{1 + x^{2}} + c_{2} \Rightarrow

I = \frac{3}{8} l n (x + \sqrt{3 + x^{2}}) + \frac{x}{4} \sqrt{3 + x^{2}} - \frac{1}{4} l n (x + \sqrt{1 + x^{2}})

- \frac{x}{4} \sqrt{1 + x^{2}} + C

Answered by ajfour last updated on 13/Nov/19

I = \int \frac{d x}{\sqrt{x^{2} + 1} + \sqrt{x^{2} + 3}}

= \frac{1}{2} \int (\sqrt{x^{2} + 3} - \sqrt{x^{2} + 1}) d x

= \frac{x}{4} (\sqrt{x^{2} + 3} - \sqrt{x^{2} + 1})

+ \frac{3}{4} \ln ∣ x + \sqrt{x^{2} + 3} ∣ - \frac{1}{4} \ln ∣ x + \sqrt{x^{2} + 1} ∣ + c

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Commented by abdomathmax last updated on 17/Nov/19

t h a n k s s i r .