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Question Number 37036 by ajfour last updated on 08/Jun/18

$${\int }_0^a(1-\frac{b-x}{\sqrt{{(b-x)}^2+cx}})dx=?$$

Commented by math khazana by abdo last updated on 09/Jun/18

$$I=a-\frac{1}{2}{\int }_0^a\frac{2(b-x)+2c}{\sqrt{{(b-x)}^2+c}}dx$$$$=a-{\int }_0^a\frac{2(b-x)+c}{2\sqrt{{(b-x)}^2+cx}}dx-\frac{c}{2}{\int }_0^a\frac{dx}{\sqrt{{(b-x)}^2+cx}}$$$$=a-\sqrt{{(b-x)}^2+cx}-\frac{c}{2}{\int }_0^a\frac{dx}{\sqrt{x^2-2bx+b^2+cx}}$$$$but{x}^2-2bx+b^2+cx=x^2+(c-2b)x+b^2$$$$=x^2+2\frac{c-2b}{2}x+\frac{{(c-2b)}^2}{4}+b^2-\frac{{(c-2b)}^2}{4}$$$$={(x+\frac{c-2b}{2})}^2+\frac{4b^2-c^2+4bc-4b^2}{4}$$$$={(x+\frac{c-2b}{2})}^2+\frac{4bc-c^2}{4}$$$$case1if4bc-c^2>0weusethechangement$$$$x+\frac{c-2b}{2}=\frac{\sqrt{4bc-c^2}}{2}tand$$$${\int }_0^a\frac{dx}{\sqrt{x^2+(c-2b)x+b^2}}={\int }_{\frac{c-2b}{\sqrt{4bc-c^2}}}^{\frac{2a+c-b}{\sqrt{4bc-c^2}}}\frac{1}{\frac{\sqrt{4bc-c^2}}{2}\sqrt{1+t^2}}\frac{\sqrt{4bc-c^2}}{2}dt$$$$={\left[argsht\right]}_{\frac{c-2b}{\sqrt{4bc-c^2}}}^{\frac{2a+c-b}{\sqrt{4bc-c^2}}}=ln(\frac{2a+c}{\sqrt{4bc-c^2}}+\sqrt{1+{\left(\frac{2a+c}{\sqrt{4bc-c^2}}\right)}^2})$$$$-ln(\frac{c-2b}{\sqrt{4bc-c^2}}+\sqrt{1+{\left(\frac{c-2b}{\sqrt{4bc-c^2}}\right)}^2})=Aso$$$$I=a-\sqrt{{(b-x)}^2+cx}-\frac{c}{2}A$$$$case24bc-c^2<0\Rightarrow {(x+\frac{c-2b}{2})}^2+\frac{4bc-c^2}{4}$$$$={(x+\frac{c-2b}{2})}^2-{\left(\sqrt{\frac{c^2-4bc}{4}}\right)}^{2}andweuse$$$$thechang.x+\frac{c-2b}{2}=\frac{\sqrt{c^2-4bc}}{2}u...becontinued...$$

Commented by ajfour last updated on 09/Jun/18

$$thankyousir,amtryingto$$$$comprehend..$$

Commented by math khazana by abdo last updated on 09/Jun/18

$$youarewelcomesirAjfour$$

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