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Question Number 95546 by Fikret last updated on 25/May/20

∫x^2 (√(a^2 +x^2 ))dx=?

$$\int{x}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}=? \\ $$

Answered by MJS last updated on 25/May/20

∫x^2 (√(x^2 +a^2 ))dx= [t=((x+(√(x^2 +a^2 )))/a) → dx=((a(√(x^2 +a^2 )))/(x+(√(x^2 +a^2 ))))dt] =(a^4 /(16))∫((t^8 −2t^4 +1)/t^5 )dt= =((a^4 (t^8 −1))/(64t^4 ))−(a^4 /8)ln t = =(1/8)x(2x^2 +a^2 )(√(x^2 +a^2 ))−(a^4 /8)ln (x+(√(x^2 +a^2 ))) +C

$$\int{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}}\:\rightarrow\:{dx}=\frac{{a}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{dt}\right] \\ $$$$=\frac{{a}^{\mathrm{4}} }{\mathrm{16}}\int\frac{{t}^{\mathrm{8}} −\mathrm{2}{t}^{\mathrm{4}} +\mathrm{1}}{{t}^{\mathrm{5}} }{dt}= \\ $$$$=\frac{{a}^{\mathrm{4}} \left({t}^{\mathrm{8}} −\mathrm{1}\right)}{\mathrm{64}{t}^{\mathrm{4}} }−\frac{{a}^{\mathrm{4}} }{\mathrm{8}}\mathrm{ln}\:{t}\:= \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}{x}\left(\mathrm{2}{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }−\frac{{a}^{\mathrm{4}} }{\mathrm{8}}\mathrm{ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)\:+\mathrm{C} \\ $$

Answered by mathmax by abdo last updated on 26/May/20

I =∫ x^2 (√(a^2 +x^2 ))dx we do the cha7gement x =asht ⇒ ∫ x^2 (√(a^2 +x^2 ))dx =∫a^2 sh^2 (t)∣a∣ch(t)ach(t)dt =a^3 ∣a∣ ∫ (sht cht)^2 dt =((a^3 ∣a∣)/4) ∫ sh^2 (2t)dt = =((a^3 ∣a∣)/4) ∫((ch(4t)−1)/2)dt =((a^3 ∣a∣)/8) ∫ ch(4t)dt −((a^3 ∣a∣)/8)t =((a^3 ∣a∣)/(32))sh(4t)−((a^3 ∣a∣)/8) t +C =((a^3 ∣a∣)/(32))(((e^(4t) −e^(−4t) )/2)) −((a^3 ∣a∣)/8)argsh((x/a))+C =((a^3 ∣a∣)/(64))((ln((x/a)+(√(1+(x^2 /a^2 ))))^4 −(ln((x/a)+(√(1+(x^2 /a^2 ))))^(−4) ) −((a^3 ∣a∣)/8)ln((x/a)+(√(1+(x^2 /a^2 )))) +C

$$\mathrm{I}\:=\int\:\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{cha7gement}\:\mathrm{x}\:=\mathrm{asht}\:\Rightarrow \\ $$$$\int\:\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{a}^{\mathrm{2}} \:+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\int\mathrm{a}^{\mathrm{2}} \:\mathrm{sh}^{\mathrm{2}} \left(\mathrm{t}\right)\mid\mathrm{a}\mid\mathrm{ch}\left(\mathrm{t}\right)\mathrm{ach}\left(\mathrm{t}\right)\mathrm{dt} \\ $$$$=\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid\:\int\:\left(\mathrm{sht}\:\mathrm{cht}\right)^{\mathrm{2}} \:\mathrm{dt}\:=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{4}}\:\int\:\mathrm{sh}^{\mathrm{2}} \left(\mathrm{2t}\right)\mathrm{dt}\:= \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{4}}\:\int\frac{\mathrm{ch}\left(\mathrm{4t}\right)−\mathrm{1}}{\mathrm{2}}\mathrm{dt}\:=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{8}}\:\int\:\mathrm{ch}\left(\mathrm{4t}\right)\mathrm{dt}\:−\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{8}}\mathrm{t} \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{32}}\mathrm{sh}\left(\mathrm{4t}\right)−\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{8}}\:\mathrm{t}\:+\mathrm{C}\: \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{32}}\left(\frac{\mathrm{e}^{\mathrm{4t}} −\mathrm{e}^{−\mathrm{4t}} }{\mathrm{2}}\right)\:−\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{8}}\mathrm{argsh}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{C} \\ $$$$=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{64}}\left(\left(\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\right)^{\mathrm{4}} −\left(\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\right)^{−\mathrm{4}} \right)\right.\right. \\ $$$$−\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{8}}\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\right)\:+\mathrm{C} \\ $$

Commented by mathmax by abdo last updated on 26/May/20

sorry I =((a^3 ∣a∣)/(64)){ ((x/a)+(√(1+(x^2 /a^2 ))))^4 −((x/a)+(√(1+(x^2 /a^2 ))))^(−4) } −((a^3 ∣a∣)/8)ln((x/a)+(√(1+(x^2 /a^2 )))) +C

$$\mathrm{sorry}\:\mathrm{I}\:=\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{64}}\left\{\:\left(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\right)^{\mathrm{4}} −\left(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\right)^{−\mathrm{4}} \right\} \\ $$$$−\frac{\mathrm{a}^{\mathrm{3}} \mid\mathrm{a}\mid}{\mathrm{8}}\mathrm{ln}\left(\frac{\mathrm{x}}{\mathrm{a}}+\sqrt{\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }}\right)\:+\mathrm{C} \\ $$

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