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Question Number 55780 by gunawan last updated on 04/Mar/19

∫ (1/(√(x^2 +2x+1))) dx= A log ∣x+1∣+C for  x>−1, then A=_____.

$$\int\:\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}\:{dx}=\:{A}\:\mathrm{log}\:\mid{x}+\mathrm{1}\mid+{C}\:\mathrm{for} \\ $$ $${x}>−\mathrm{1},\:\mathrm{then}\:{A}=\_\_\_\_\_. \\ $$

Commented bymaxmathsup by imad last updated on 04/Mar/19

∫ (dx/(√(x^2 +2x+1))) =∫  (dx/(∣x+1∣)) =∫(dx/(x+1))  if x>−1 ⇒  ∫  (dx/(√(x^2 +2x+1))) =ln∣x+1∣ +c ⇒A =1 .

$$\int\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}\:=\int\:\:\frac{{dx}}{\mid{x}+\mathrm{1}\mid}\:=\int\frac{{dx}}{{x}+\mathrm{1}}\:\:{if}\:{x}>−\mathrm{1}\:\Rightarrow \\ $$ $$\int\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}}\:={ln}\mid{x}+\mathrm{1}\mid\:+{c}\:\Rightarrow{A}\:=\mathrm{1}\:. \\ $$

Answered by MJS last updated on 04/Mar/19

A=1  (√(x^2 +2x+1))=∣x+1∣

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

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