calculate ∫ ((x+1−(√(2x+3)))/(x−2 +(√(x+1)))) dx


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Question Number 98942 by mathmax by abdo last updated on 17/Jun/20

$calculate \int \frac{x + 1 - \sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} dx$
Commented by MJS last updated on 17/Jun/20

$Sir Abdo, do you have an easy solution ?$ $I can solve it but the constants are getting$ $horrible . . .$
Commented by mathmax by abdo last updated on 18/Jun/20

$use hyperbolic function sir .$
	Answered by 4635 last updated on 17/Jun/20

	$\Leftrightarrow \int \frac{x + 1}{x - 2 + \sqrt{x + 1}} d x - \int \frac{\sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} d x$
	Commented by 4635 last updated on 17/Jun/20

	$p o s o n s k = \int \frac{x + 1}{x - 2 + \sqrt{x + 1}} d x e t$ $d = \int \frac{\sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} d x$
	Commented by 4635 last updated on 17/Jun/20

	$o n a : \int \frac{x + 1}{x - 2 + \sqrt{x + 1}} d x$ $p u t t^{2} = x + 1 \Rightarrow 2 t d t = d x$ $\int \frac{2 t^{3}}{t^{2} - 3 + t} d t \Rightarrow \int \frac{2 t^{3}}{t^{2} + t - 3} d t$ $= \int \frac{2 t^{3}}{{(t + \frac{1}{2})}^{2} - {(\frac{\sqrt{7}}{2})}^{2}} d t$ $l e t u = t + \frac{1}{2} \Rightarrow d u = d t$ $\Rightarrow 2 \int \frac{{(u - \frac{1}{2})}^{3}}{u^{2} - {(\frac{\sqrt{7}}{2})}^{2}} d u = \int \frac{u^{3} - \frac{3 u^{2}}{2} + \frac{3 u}{4} - \frac{1}{8}}{u^{2} - {(\frac{\sqrt{7}}{2})}^{2}} d u$
	Answered by MJS last updated on 17/Jun/20

	$\int \frac{x + 1 - \sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} d x =$ $= \int \frac{x + 1}{x - 2 + \sqrt{x + 1}} d x - \int \frac{\sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} d x$ $\int \frac{x + 1}{x - 2 + \sqrt{x + 1}} d x =$ $[t = \sqrt{x + 1} \to d x = 2 \sqrt{x + 1} d t]$ $= 2 \int \frac{t^{3}}{t^{2} + t - 3} d t = 2 (\int t d t - \int d t + \int \frac{4 t - 3}{t^{2} + t - 3} d t) =$ $= t (t - 2) + \frac{4}{13} ((26 + 5 \sqrt{13}) \int \frac{d t}{2 t + 1 + \sqrt{13}} + (26 - 5 \sqrt{13}) \int \frac{d t}{2 t + 1 - \sqrt{13}})$ $and these are easy to solve$ $- \int \frac{\sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} d x = - \int \frac{(x - 2 - \sqrt{x + 1}) \sqrt{2 x + 3}}{x^{2} - 5 x + 3} d x =$ $= - \int \frac{(x - 2) \sqrt{2 x + 3}}{x^{2} - 5 x + 3} d x + \int \frac{\sqrt{x + 1} \sqrt{2 x + 3}}{x^{2} - 5 x + 3} d x$ $- \int \frac{(x - 2) \sqrt{2 x + 3}}{x^{2} - 5 x + 3} d x =$ $[u = \sqrt{2 x + 3} \to d x = \sqrt{2 x + 3} d u]$ $= - 2 \int \frac{u^{2} (u^{2} - 7)}{u^{4} - 16 u^{2} + 51} d u$ $u^{4} - 16 u^{2} + 51 = (u - α) (u + α) (u - β) (u + β)$ $α, β \in R$ $\Rightarrow we can easily solve this by decomposing$ $\int \frac{\sqrt{x + 1} \sqrt{2 x + 3}}{x^{2} - 5 x + 3} d x =$ $[v = \frac{\sqrt{2 (x + 1)}}{\sqrt{2 x + 3}} \to d x = \sqrt{2 (x + 1) {(2 x + 3)}^{3}} d v]$ $= 2 \sqrt{2} \int \frac{v^{2}}{(v^{2} - 1) (51 v^{4} - 86 v^{2} + 36)} d v$ $51 v^{4} - 86 v^{2} + 36 = 51 (v - γ) (v + γ) (v - δ) (v + γ)$ $γ, δ \in R$ $\Rightarrow again easy to solve$ $. . . but I^{'} m not eager for typing the constants . . .$
	Commented by mathmax by abdo last updated on 18/Jun/20

	$thank you sir mjs .$
	Answered by mathmax by abdo last updated on 18/Jun/20

	$I = \int \frac{x + 1 - \sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} dx \Rightarrow I = \int \frac{x + 1}{x - 2 + \sqrt{x + 1}} dx - \int \frac{\sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} dx = H - K$ $changement \sqrt{x + 1} = t give x + 1 = t^{2} \Rightarrow$ $H = \int \frac{t^{2}}{t^{2} - 1 - 2 + t} (2 t) dt = 2 \int \frac{t^{3}}{t^{2} + t - 3} dt$ $= 2 \int \frac{t (t^{2} + t - 3) - t^{2} + 3 t}{t^{2} + t - 3} dt = 2 \int tdt + 2 \int \frac{- t^{2} + 3 t}{t^{2} + t - 3} dt$ $= t^{2} - 2 \int \frac{t^{2} - 3 t}{t^{2} + t - 3} dt = t^{2} - 2 \int \frac{t^{2} + t - 3 - t + 3 - 3 t}{t^{2} + t - 3} dt = t^{2} - 2 t + 2 \int \frac{4 t - 3}{t^{2} + t - 3} dt$ $t^{2} + t - 3 = 0 \to Δ = 1 + 12 = 13 \Rightarrow t_{1} = \frac{- 1 + \sqrt{13}}{2} and t_{2} = \frac{- 1 - \sqrt{13}}{2}$ $f (t) = \frac{4 t - 3}{t^{2} + t - 3} = \frac{4 t - 3}{(t - t_{1}) (t - t_{2})} = \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} \Rightarrow eazy to find a and b \Rightarrow$ $\int f (t) dt = aln ∣ t - t_{1} ∣ + bln ∣ t - t_{2} ∣ + c \Rightarrow$ $H = x + 1 - 2 \sqrt{x + 1} + 2 aln ∣ \sqrt{x + 1} - t_{1} ∣ + 2 b \ln ∣ \sqrt{x + 1} - t_{2} ∣ + c$ $K = \int \frac{\sqrt{2 x + 3}}{x - 2 + \sqrt{x + 1}} dt changement \sqrt{x + 1} = t give x + 1 = t^{2}$ $K = \int \frac{\sqrt{2 (t^{2} - 1) + 3}}{t^{2} - 1 - 2 + t} (2 t) dt = 2 \int \frac{\sqrt{{2 t}^{2} + 1}}{t^{2} + t - 3} tdt$ $= 2 \int \frac{t \sqrt{2 (t^{2} + \frac{1}{2})}}{t^{2} + t - 3} dt = 2 \sqrt{2} \int \frac{t \sqrt{t^{2} + \frac{1}{2}}}{t^{2} + t - 3} dt we do the changement$ $t = \frac{1}{\sqrt{2}} shu \Rightarrow K = \sqrt{2} \int \frac{shu \times chu}{\frac{1}{2} {sh}^{2} u + \frac{1}{\sqrt{2}} shu - 3} (\frac{1}{\sqrt{2}}) chu du$ $= \int \frac{2 shu {ch}^{2} u}{{sh}^{2} u + \sqrt{2} shu - 6} du this integrale is solvable . . . be continued . . .$
	Commented by MJS last updated on 18/Jun/20

	$this leads to a polynome of 4^{th} degree in the$ $denominator, too . we can solve it but the$ $constants are horrible also$
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