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 bymaxmathsup 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 | ||
$${A}=\mathrm{1} \\ $$ $$\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}=\mid{x}+\mathrm{1}\mid \\ $$ | ||