Question Number 166346 by LEKOUMA last updated on 18/Feb/22
$$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}} \\ $$
Commented by mkam last updated on 18/Feb/22
$$\boldsymbol{{w}}\:=\:\mathrm{1}\:+\:\sqrt{\boldsymbol{{x}}}\:+\sqrt{\mathrm{1}+\boldsymbol{{x}}}\:\Rightarrow\:\boldsymbol{{dw}}\:=\:\left(\frac{\mathrm{1}}{\mathrm{2}\sqrt{\boldsymbol{{x}}}\:+\mathrm{2}\sqrt{\mathrm{1}+\boldsymbol{{x}}}}\right)\boldsymbol{{dx}} \\ $$$$ \\ $$$$\Rightarrow\:\boldsymbol{{dw}}\:=\:\left(\:\frac{\mathrm{1}}{\mathrm{2}\:\boldsymbol{{w}}\:−\:\mathrm{2}}\:\right)\:\boldsymbol{{dx}}\:\Rightarrow\:\boldsymbol{{dw}}\:\left(\:\mathrm{2}\:\boldsymbol{{w}}\:−\:\mathrm{2}\:\right)\:=\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\int\:\:\frac{\mathrm{2}\:\boldsymbol{{w}}\:−\:\mathrm{2}}{\boldsymbol{{w}}}\:\boldsymbol{{dw}}\:=\:\int\:\left(\:\mathrm{2}\:−\:\frac{\mathrm{2}}{\boldsymbol{{w}}}\:\right)\:\boldsymbol{{dw}}\:=\:\mathrm{2}\boldsymbol{{w}}\:−\:\mathrm{2}\:\boldsymbol{{ln}}\mid\:\boldsymbol{{w}}\:\mid\:+\:\boldsymbol{{C}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\mathrm{2}\:+\:\mathrm{2}\:\sqrt{\boldsymbol{{x}}}\:+\:\mathrm{2}\:\sqrt{\mathrm{1}+\boldsymbol{{x}}}\:−\:\mathrm{2}\:\boldsymbol{{ln}}\:\mid\:\mathrm{1}\:+\:\sqrt{\boldsymbol{{x}}}\:+\:\sqrt{\mathrm{1}+\boldsymbol{{x}}}\:\mid\:+\:\boldsymbol{{C}} \\ $$$$ \\ $$$$\square\:\boldsymbol{{mohammad}} \\ $$
Answered by MJS_new last updated on 18/Feb/22
$$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+\sqrt{{x}}−\sqrt{\mathrm{1}+{x}}}{\:\sqrt{{x}}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}−\frac{\sqrt{{x}+\mathrm{1}}}{\:\sqrt{{x}}}\right){dx}= \\ $$$$… \\ $$$$=\frac{{x}}{\mathrm{2}}+\sqrt{{x}}−\frac{\sqrt{{x}}\sqrt{{x}+\mathrm{1}}}{\mathrm{2}}−\frac{\mathrm{ln}\:\left(\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\right)}{\mathrm{2}}+{C} \\ $$