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Question-227271




Question Number 227271 by Spillover last updated on 11/Jan/26
Answered by Kassista last updated on 11/Jan/26
  ∫ (x^2 /(x^2 +5)) dx = ∫ ((x^2 +5−5)/(x^2 +5)) dx = ∫ ((x^2 +5)/(x^2 +5)) dx −∫(5/(x^2 +5)) dx   ⇒^(x=(√5)tan θ)  x −5 ∫ (((√5)sec^2 θ)/( 5sec^2  θ)) dθ = x−(√5)∫ dθ   = x −(√5)arctan((x/( (√5))))+C
$$ \\ $$$$\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\:=\:\int\:\frac{{x}^{\mathrm{2}} +\mathrm{5}−\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\:=\:\int\:\frac{{x}^{\mathrm{2}} +\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\:−\int\frac{\mathrm{5}}{{x}^{\mathrm{2}} +\mathrm{5}}\:{dx}\: \\ $$$$\overset{{x}=\sqrt{\mathrm{5}}\mathrm{tan}\:\theta} {\Rightarrow}\:{x}\:−\mathrm{5}\:\int\:\frac{\sqrt{\mathrm{5}}\mathrm{sec}\:^{\mathrm{2}} \theta}{\:\mathrm{5sec}^{\mathrm{2}} \:\theta}\:{d}\theta\:=\:{x}−\sqrt{\mathrm{5}}\int\:{d}\theta\: \\ $$$$=\:{x}\:−\sqrt{\mathrm{5}}{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{5}}}\right)+{C} \\ $$
Answered by Spillover last updated on 11/Jan/26
Answered by Spillover last updated on 11/Jan/26

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