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Question Number 20293 by tammi last updated on 25/Aug/17

∫(√(((a+x)/x)dx))

$$\int\sqrt{\frac{{a}+{x}}{{x}}{dx}} \\ $$

Answered by $@ty@m last updated on 25/Aug/17

=∫((a+x)/(√(x(a+x))))dx =∫((a+x)/(√(ax+x^2 )))dx =(1/2)∫((2x+a+a)/(√(ax+x^2 )))dx =(1/2)∫((2x+a)/(√(ax+x^2 )))dx+(a/2)∫(dx/(√(ax+x^2 ))) =(√(ax+x^2 ))+(a/2)∫(dx/(√(x^2 +2.(a/2)x+((a/2))^2 −((a/2))^2 )))dx =(√(ax+x^2 ))+(a/2)∫(dx/(√((x+(a/2))^2 −((a/2))^2 ))) =(√(ax+x^2 ))+(a/2)ln∣x+(a/2)+(√(ax+x^2 ))∣+C

$$=\int\frac{{a}+{x}}{\sqrt{{x}\left({a}+{x}\right)}}{dx} \\ $$$$=\int\frac{{a}+{x}}{\sqrt{{ax}+{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{a}+{a}}{\sqrt{{ax}+{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+{a}}{\sqrt{{ax}+{x}^{\mathrm{2}} }}{dx}+\frac{{a}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{{ax}+{x}^{\mathrm{2}} }} \\ $$$$=\sqrt{{ax}+{x}^{\mathrm{2}} }+\frac{{a}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}.\frac{{a}}{\mathrm{2}}{x}+\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}{dx} \\ $$$$=\sqrt{{ax}+{x}^{\mathrm{2}} }+\frac{{a}}{\mathrm{2}}\int\frac{{dx}}{\sqrt{\left({x}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\sqrt{{ax}+{x}^{\mathrm{2}} }+\frac{{a}}{\mathrm{2}}{ln}\mid{x}+\frac{{a}}{\mathrm{2}}+\sqrt{{ax}+{x}^{\mathrm{2}} }\mid+{C} \\ $$

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