Question Number 55780 by gunawan last updated on 04/Mar/19
$$\int\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}\:{dx}=\:{A}\:\mathrm{log}\:\mid{x}+\mathrm{1}\mid+{C}\:\mathrm{for} \\ $$$${x}>−\mathrm{1},\:\mathrm{then}\:{A}=\_\_\_\_\_. \\ $$
Commented by maxmathsup by imad last updated on 04/Mar/19
$$\int\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}\:=\int\:\:\frac{{dx}}{\mid{x}+\mathrm{1}\mid}\:=\int\frac{{dx}}{{x}+\mathrm{1}}\:\:{if}\:{x}>−\mathrm{1}\:\Rightarrow \\ $$$$\int\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}\:={ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow{A}\:=\mathrm{1}\:. \\ $$
Answered by MJS last updated on 04/Mar/19
$${A}=\mathrm{1} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}=\mid{x}+\mathrm{1}\mid \\ $$