Previous in Relation and Functions Next in Relation and Functions
Question Number 148812 by mathmax by abdo last updated on 31/Jul/21
$$\mathrm{find}\:\int\:\:\frac{\mathrm{dx}}{\left(\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}+\mathrm{1}}\right)\left(\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{x}}\right)} \\ $$
Answered by MJS_new last updated on 31/Jul/21
![∫(dx/(((√x)+(√(x+1)))((√x)+(√(x−1)))))= =∫((1/(((√x)+(√(x+1)))((√x)+(√(x−1)))))×((((√x)−(√(x+1)))((√x)−(√(x−1))))/(((√x)−(√(x+1)))((√x)−(√(x−1))))))dx= =∫(x+(√(x^2 −1))−(√(x(x−1)))−(√(x(x+1))))dx [which is easy to solve] =(x^2 /2)+((x(√(x^2 −1))−ln (x+(√(x^2 −1))))/2)−((2(2x−1)(√(x(x−1)))−ln (2x−1+2(√(x(x−1)))))/8)−((2(2x+1)(√(x(x+1)))−ln (2x+1+2(√(x(x+1)))))/8)+C](Q148826.png)
$$\int\frac{{dx}}{\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)\left(\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\right)}= \\ $$$$=\int\left(\frac{\mathrm{1}}{\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)\left(\sqrt{{x}}+\sqrt{{x}−\mathrm{1}}\right)}×\frac{\left(\sqrt{{x}}−\sqrt{{x}+\mathrm{1}}\right)\left(\sqrt{{x}}−\sqrt{{x}−\mathrm{1}}\right)}{\left(\sqrt{{x}}−\sqrt{{x}+\mathrm{1}}\right)\left(\sqrt{{x}}−\sqrt{{x}−\mathrm{1}}\right)}\right){dx}= \\ $$$$=\int\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\sqrt{{x}\left({x}−\mathrm{1}\right)}−\sqrt{{x}\left({x}+\mathrm{1}\right)}\right){dx} \\ $$$$\:\:\:\:\:\left[\mathrm{which}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\right] \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)}{\mathrm{2}}−\frac{\mathrm{2}\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}\left({x}−\mathrm{1}\right)}−\mathrm{ln}\:\left(\mathrm{2}{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}\left({x}−\mathrm{1}\right)}\right)}{\mathrm{8}}−\frac{\mathrm{2}\left(\mathrm{2}{x}+\mathrm{1}\right)\sqrt{{x}\left({x}+\mathrm{1}\right)}−\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)}\right)}{\mathrm{8}}+{C} \\ $$
Commented by MJS_new last updated on 01/Aug/21
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}\:\mathrm{as}\:\mathrm{always}\:\mathrm{and}\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{your} \\ $$$$\left.\mathrm{concern};\:\mathrm{I}'\mathrm{m}\:\mathrm{well}.\:\mathrm{Maybe}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{worn}\:\mathrm{out}\:;−\right) \\ $$
Commented by mathmax by abdo last updated on 01/Aug/21
$$\mathrm{thank}\:\mathrm{sir}\:\mathrm{mjs}\:\mathrm{you}\:\mathrm{are}\:\mathrm{absent}\:\mathrm{for}\:\mathrm{a}\:\mathrm{time}\:\mathrm{i}\:\mathrm{hope}\:\mathrm{that}\:\mathrm{your}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{good}\:\mathrm{helth}.... \\ $$