Question Number 128346 by Ahmed1hamouda last updated on 06/Jan/21
Answered by MJS_new last updated on 06/Jan/21
![∫x^2 (√(4x^2 +1))dx= [t=2x+(√(4x^2 +1)) → dx=((√(4x^2 +1))/(2(2x+(√(4x^2 +1)))))dt] =∫((t^3 /(128))−(1/(64t))+(1/(128t^5 )))dt= =(t^4 /(512))−((ln t)/(64))−(1/(512t^4 ))= =((x(8x^2 +1)(√(4x^2 +1)))/(32))−(1/(64))ln (2x+(√(4x^2 +1))) +C ⇒ answer is ((9(√5))/(32))−(1/(64))ln (2+(√5))](Q128356.png)
$$\int{x}^{\mathrm{2}} \sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}\left(\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\right)}{dt}\right] \\ $$$$=\int\left(\frac{{t}^{\mathrm{3}} }{\mathrm{128}}−\frac{\mathrm{1}}{\mathrm{64}{t}}+\frac{\mathrm{1}}{\mathrm{128}{t}^{\mathrm{5}} }\right){dt}= \\ $$$$=\frac{{t}^{\mathrm{4}} }{\mathrm{512}}−\frac{\mathrm{ln}\:{t}}{\mathrm{64}}−\frac{\mathrm{1}}{\mathrm{512}{t}^{\mathrm{4}} }= \\ $$$$=\frac{{x}\left(\mathrm{8}{x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{ln}\:\left(\mathrm{2}{x}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\right)\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{9}\sqrt{\mathrm{5}}}{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{ln}\:\left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$
Answered by mathmax by abdo last updated on 06/Jan/21
![I=∫_0 ^1 (x^2 (√(4x^2 +1)))dx we do the changement 2x=sh(t) ⇒ I =∫_0 ^(argsh(2)) (((sht)/2))^2 cht ×((chtdt)/2) =(1/8)∫_0 ^(ln(2+(√(1+2^2 ))) ) sh^2 t ch^2 t dt =(1/8)∫_0 ^(ln(2+(√5))) (((sh(2t))/2))^2 dt =(1/(32))∫_0 ^(ln(2+(√5))) ((ch(4t)−1)/2)dt =(1/(64))∫_0 ^(ln(2+(√5))) (ch(4t)−1)dt =(1/(4.64))[ch(4t)]_0 ^(ln(2+(√5))) −((ln(2+(√5)))/(64)) =(1/(4.64))[((e^(4t) +e^(−4t) )/2)]_0 ^(ln(2+(√5))) −((ln(2+(√5)))/(64)) =(1/(8.64)){(2+(√5))^4 +(2+(√5))^(−4) −2}](Q128389.png)
$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{2x}=\mathrm{sh}\left(\mathrm{t}\right)\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{argsh}\left(\mathrm{2}\right)} \left(\frac{\mathrm{sht}}{\mathrm{2}}\right)^{\mathrm{2}} \mathrm{cht}\:×\frac{\mathrm{chtdt}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{1}+\mathrm{2}^{\mathrm{2}} }\right)\:} \mathrm{sh}^{\mathrm{2}} \mathrm{t}\:\mathrm{ch}^{\mathrm{2}} \:\mathrm{t}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \left(\frac{\mathrm{sh}\left(\mathrm{2t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{32}}\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \frac{\mathrm{ch}\left(\mathrm{4t}\right)−\mathrm{1}}{\mathrm{2}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\int_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \left(\mathrm{ch}\left(\mathrm{4t}\right)−\mathrm{1}\right)\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{4}.\mathrm{64}}\left[\mathrm{ch}\left(\mathrm{4t}\right)\right]_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} −\frac{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{64}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}.\mathrm{64}}\left[\frac{\mathrm{e}^{\mathrm{4t}} +\mathrm{e}^{−\mathrm{4t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} −\frac{\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)}{\mathrm{64}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}.\mathrm{64}}\left\{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{4}} +\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{−\mathrm{4}} −\mathrm{2}\right\} \\ $$