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Question Number 125053 by bramlexs22 last updated on 08/Dec/20

$$\int \frac{dx}{\sqrt{x^2+3x-4}}=?$$

Commented by Dwaipayan Shikari last updated on 08/Dec/20

$$\int \frac{dx}{\sqrt{{(x+\frac{3}{2})}^2-{\left(\frac{5}{2}\right)}^2}}x+\frac{3}{2}=\frac{5}{2}sec\theta \Rightarrow =\frac{5}{2}sec\theta tan\theta \frac{d\theta }{dx}$$$$\frac{5}{2}\int \frac{sec\theta tan\theta d\theta }{tan\theta }=\frac{5}{2}log(sec\theta +tan\theta )=\frac{5}{2}log(\frac{2x+3}{5}+\frac{\sqrt{4x^2+12x-16}}{5})+C$$

Answered by benjo_mathlover last updated on 08/Dec/20

$$let\sqrt{x^2+3x-4}=(x+4).t$$$$x-1=(x+4)t^2\Rightarrow x=\frac{1+4t^2}{1-t^2}$$$$dx=\frac{10t}{{(1-t^2)}^2}dt\wedge \sqrt{(x+4)(x-1)}=\frac{5t}{1-t^2}$$$$I=\int \frac{2}{1-t^2}dt=\ell n\mid \frac{1+t}{1-t}\mid +c$$$$I=\ell n\mid \frac{1+\sqrt{\frac{x-1}{x+4}}}{1-\sqrt{\frac{x-1}{x+4}}}\mid +c$$$$I=\ell n\mid \frac{\sqrt{x+4}+\sqrt{x-1}}{\sqrt{x+4}-\sqrt{x-1}}\mid +c$$

Commented by bramlexs22 last updated on 08/Dec/20

$$EulerSubstitutionmethod$$

Answered by Bird last updated on 08/Dec/20

$$I=\int \frac{dx}{\sqrt{x^2+3x-4}}$$$$x^2+3x-4=0\to \mathrm{\Delta }=9+16=25$$$$\Rightarrow x_1=\frac{-3+5}{2}=1and{x}_2=\frac{-3-5}{2}=-4$$$$\Rightarrow I=\int \frac{dx}{\sqrt{(x-1)(x+4)}}$$$$wedothechangement\sqrt{x-1}=t$$$$\Rightarrow x-1=t^2\Rightarrow I=\int \frac{2tdt}{t\sqrt{t^2+1+4}}$$$$=2\int \frac{dt}{\sqrt{t^2+5}}{=}_{t=\sqrt{5}u}2\int \frac{\sqrt{5}du}{\sqrt{5}\sqrt{1+u^2}}$$$$=2\int \frac{du}{\sqrt{1+u^2}}=2ln(u+\sqrt{1+u^2})+c$$$$=2ln(\frac{t}{\sqrt{5}}+\sqrt{1+\frac{t^2}{5}})+c$$$$=2ln(\frac{\sqrt{x-1}}{\sqrt{5}}+\sqrt{1+\frac{x-1}{5}})+C$$$$$$

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