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Question Number 48719 by Abdo msup. last updated on 27/Nov/18

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

Commented by Abdo msup. last updated on 27/Nov/18

$$I=\int \frac{x-2}{\sqrt{x^2+4x+4-7}}=\int \frac{x-2}{\sqrt{{(x+2)}^2-7}}dx$$$${=}_{x+2=\sqrt{7}ch\left(t\right)}\int \frac{\sqrt{7}ch\left(t\right)-4}{\sqrt{7}sh\left(t\right)}\sqrt{7}sh\left(t\right)dt$$$$=\sqrt{7}\int ch\left(t\right)-4t+c=\sqrt{7}sh\left(t\right)-4t+cbut$$$$sh\left(t\right)=\frac{e^t-e^{-t}}{2}andt=argch\left(\frac{x+2}{\sqrt{7}}\right)$$$$=ln(x+\sqrt{{\left(\frac{x+2}{\sqrt{7}}\right)}^2-1})\Rightarrow sh\left(t\right)=\frac{x+\sqrt{{\left(\frac{x+2}{\sqrt{7}}\right)}^2-1}-\frac{1}{x+\sqrt{\left(\frac{x+2}{\sqrt{7}}\right)-1}}}{2}$$$$I=\frac{\sqrt{7}}{2}\{(x+\sqrt{{\left(\frac{x+2}{\sqrt{7}}\right)}^2-1}-{(x+\sqrt{{\left(\frac{x+2}{\sqrt{7}}\right)}^2-1})}^{-1}\}-4t+c.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Nov/18

$$t^2=x^2+4x-3$$$$2t\frac{dt}{dx}=2x+4$$$$tdt=(x+2)dx$$$$\int \frac{x+2-4}{\sqrt{x^2+4x-3}}dx$$$$\int \frac{tdt}{t}-4\int \frac{dx}{\sqrt{{(x+2)}^2-{\left(\sqrt{7}\right)}^2}}$$$$t-4ln\{(x+2)+\sqrt{{(x+2)}^2-{\left(\sqrt{7}\right)}^2}\}+c$$$$\sqrt{(x^2+4x-3)}-4ln\{(x+2)+\sqrt{x^2+4x-3}\}+c$$

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