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Question Number 33145 by Ahmad Hajjaj last updated on 11/Apr/18

$$\underset{0}{\stackrel{3}{\int }}\frac{dx}{\sqrt{x+1}+\sqrt{5x+1}}=$$

Commented by prof Abdo imad last updated on 11/Apr/18

$$I={\int }_0^3\frac{dx}{\sqrt{x+1}+\sqrt{5x+1}}.ch.\sqrt{x+1}=t\Rightarrow$$$$I={\int }_1^2\frac{2tdt}{t+\sqrt{5(t^2-1)+1}}$$$$={\int }_1^2\frac{2tdt}{t+\sqrt{5t^2-4}}={\int }_1^2\frac{2dt}{1+\sqrt{\frac{5t^2-4}{t^2}}}$$$$={\int }_1^2\frac{2dt}{1+\sqrt{5-\frac{4}{t^2}}}afterweusethech.$$$$\frac{2}{t}=\sqrt{5}sinx\Rightarrow \frac{t}{2}=\frac{1}{\sqrt{5}sinx}\Rightarrow t=\frac{2}{\sqrt{5}sinx}$$$$dt=-\frac{2}{\sqrt{5}}\frac{cosx}{{sin}^{2}x}andx=arcsin\left(\frac{2}{t\sqrt{5}}\right)$$$$I={\int }_{arcsin\left(\frac{2}{\sqrt{5}}\right)}^{arcsin\left(\frac{1}{\sqrt{5}}\right)}\frac{1}{1+\sqrt{5}\sqrt{1-{sin}^{2}x}}-\frac{2}{\sqrt{5}}\frac{cosx}{{sin}^{2}x}dx$$$$=-\frac{2}{\sqrt{5}}{\int }_{arcsin\left(\frac{2}{\sqrt{5}}\right)}^{arcsin\left(\frac{1}{\sqrt{5}}\right)}\frac{cosx}{{sin}^{2}x(1+\sqrt{5}cosx)}dx$$$${\int }_{\alpha }^{\beta }-\frac{cosx}{{sin}^{2}x(1+\sqrt{5}cosx)}\left(byparts\right)$$$$={\left[\frac{1}{sinx}\frac{1}{1+\sqrt{5}cosx}\right]}_{\alpha }^{\beta }-{\int }_{\alpha }^{\beta }\frac{1}{sinx}\frac{\sqrt{5}sinx}{{(1+\sqrt{5}cosx)}^2}dx$$$$=\frac{1}{sin\beta (1+\sqrt{5}cos\beta )}-\frac{1}{sin\alpha (1+\sqrt{5}cos\alpha )}$$$$-\sqrt{5}{\int }_{\alpha }^{\beta }\frac{dx}{{(1+\sqrt{5}cosx)}^2}andthisintegralis$$$$calculablebych.tan\left(\frac{x}{2}\right)=u....$$$$\alpha =arcsin\left(\frac{2}{\sqrt{5}}\right)and\beta =arcsin\left(\frac{1}{\sqrt{5}}\right).$$$$$$

Answered by MJS last updated on 11/Apr/18

$$\frac{1}{a+b}=\frac{a-b}{(a+b)(a-b)}=\frac{a-b}{a^2-b^2}$$$$$$$$\frac{1}{\sqrt{x+1}+\sqrt{5x+1}}=\frac{\sqrt{x+1}-\sqrt{5x+1}}{x+1-5x-1}=$$$$=\frac{\sqrt{5x+1}-\sqrt{x+1}}{4x}$$$$\underset{0}{\stackrel{3}{\int }}\frac{dx}{\sqrt{x+1}+\sqrt{5x+1}}=\frac{1}{4}{\left[(\int \frac{\sqrt{5x+1}}{x}dx-\int \frac{\sqrt{x+1}}{x}dx)\right]}_0^3=$$$$$$$$(\int \frac{\sqrt{ax+b}}{x}dx\mathrm{with}b>0)=$$$$=2\sqrt{ax+b}+\sqrt{b}\times \ln \mid \frac{\sqrt{ax+b}-\sqrt{b}}{\sqrt{ax+b}+\sqrt{b}}\mid$$$$$$$$=\frac{1}{4}{\left[((2\sqrt{5x+1}+\ln \mid \frac{\sqrt{5x+1}-1}{\sqrt{5x+1}+1}\mid )-(2\sqrt{x+1}+\ln \mid \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\mid ))\right]}_0^3=$$$$=\frac{1}{4}{[2(\sqrt{5x+1}-\sqrt{x+1})+\ln (\sqrt{5x+1}-1)-\ln (\sqrt{5x+1}+1)-\ln (\sqrt{x+1}-1)+\ln (\sqrt{x+1}+1)]}_0^3=$$$$=\frac{1}{4}{[2(\sqrt{5x+1}-\sqrt{x+1})+\ln \frac{\sqrt{x+1}+1}{\sqrt{5x+1}+1}+\ln \frac{\sqrt{5x+1}-1}{\sqrt{x+1}-1}]}_0^3=$$$$\underset{x\to 0}{\lim }\frac{\sqrt{5x+1}-1}{\sqrt{x+1}-1}=5$$$$=1+\frac{1}{2}\ln \frac{3}{5}$$

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