Question Number 92447 by mathmax by abdo last updated on 07/May/20
$${find}\:\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{3}{x}}+\sqrt{\mathrm{1}+\mathrm{3}{x}}} \\ $$
Commented by mathmax by abdo last updated on 07/May/20
![A =∫_(1/6) ^(1/5) (dx/( (√(1−3x))+(√(1+3x)))) ⇒A =∫_(1/6) ^(1/5) (((√(1+3x))−(√(1−3x)))/(6x))dx 6A =∫_(1/6) ^(1/5) ((√(1+3x))/x)dx−∫_(1/6) ^(1/5) ((√(1−3x))/x)dx changement (√(1+3x))=t give 1+3x=t^2 ⇒x =((t^2 −1)/3) ⇒∫_(1/6) ^(1/5) ((√(1+3x))/x)dx =3∫_(√(3/2)) ^(√(8/(5 ))) (t/(t^2 −1))×(2/3)t dt =2∫_(√(3/2)) ^(√(8/5)) ((t^2 −1 +1)/(t^2 −1))dt =2{(√(8/5))−(√(3/2))}+∫_(√(3/2)) ^(√(8/5)) ((1/(t−1))−(1/(t+1)))dt =2((√(8/5))−(√(3/2)))+[ln∣((t−1)/(t+1))∣_(√(3/2)) ^(√(8/5)) ] =2((√(8/5))−(√(3/2)))+ln∣(((√(8/5))−1)/( (√(8/5))+1))∣−ln∣(((√(3/2))−1)/( (√(3/2))+1))∣ also chang.(√(1−3x))=t give 1−3x =t^2 ⇒x=((1−t^2 )/3) ∫_(1/6) ^(1/5) ((√(1−3x))/x)dx =3∫_(√(1/2)) ^(√(2/5)) (t/(1−t^2 ))(−(2/3)t)dt =2 ∫_(1/( (√2))) ^(√(2/5)) ((t^2 −1+1)/(t^2 −1)) dt =2((√(2/5))−(1/( (√2))))+∫_(1/( (√2))) ^(√(2/5)) ((1/(t−1))−(1/(t+1)))dt =2((√(2/5))−(1/( (√2))))+[ln∣((t−1)/(t+1))∣]_(1/( (√2))) ^(√(2/5)) ?=....so A is known](https://www.tinkutara.com/question/Q92528.png)
$${A}\:=\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\frac{{dx}}{\:\sqrt{\mathrm{1}−\mathrm{3}{x}}+\sqrt{\mathrm{1}+\mathrm{3}{x}}}\:\Rightarrow{A}\:=\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\frac{\sqrt{\mathrm{1}+\mathrm{3}{x}}−\sqrt{\mathrm{1}−\mathrm{3}{x}}}{\mathrm{6}{x}}{dx} \\ $$$$\mathrm{6}{A}\:=\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\frac{\sqrt{\mathrm{1}+\mathrm{3}{x}}}{{x}}{dx}−\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\frac{\sqrt{\mathrm{1}−\mathrm{3}{x}}}{{x}}{dx}\:\:{changement}\:\sqrt{\mathrm{1}+\mathrm{3}{x}}={t}\:{give} \\ $$$$\mathrm{1}+\mathrm{3}{x}={t}^{\mathrm{2}} \:\Rightarrow{x}\:=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{3}}\:\Rightarrow\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\frac{\sqrt{\mathrm{1}+\mathrm{3}{x}}}{{x}}{dx}\:=\mathrm{3}\int_{\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{8}}{\mathrm{5}\:}}} \:\:\:\:\frac{{t}}{{t}^{\mathrm{2}} −\mathrm{1}}×\frac{\mathrm{2}}{\mathrm{3}}{t}\:{dt} \\ $$$$=\mathrm{2}\int_{\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}} \:\:\:\frac{{t}^{\mathrm{2}} −\mathrm{1}\:+\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:\:=\mathrm{2}\left\{\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}−\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right\}+\int_{\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}} \:\:\:\:\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\mathrm{2}\left(\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}−\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right)+\left[{ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid_{\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}} \:\:\right] \\ $$$$=\mathrm{2}\left(\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}−\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\right)+{ln}\mid\frac{\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{8}}{\mathrm{5}}}+\mathrm{1}}\mid−{ln}\mid\frac{\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}+\mathrm{1}}\mid \\ $$$${also}\:{chang}.\sqrt{\mathrm{1}−\mathrm{3}{x}}={t}\:{give}\:\mathrm{1}−\mathrm{3}{x}\:={t}^{\mathrm{2}} \:\Rightarrow{x}=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\frac{\sqrt{\mathrm{1}−\mathrm{3}{x}}}{{x}}{dx}\:=\mathrm{3}\int_{\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}} \:\:\:\frac{{t}}{\mathrm{1}−{t}^{\mathrm{2}} }\left(−\frac{\mathrm{2}}{\mathrm{3}}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}} \:\:\:\frac{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{t}^{\mathrm{2}} −\mathrm{1}}\:{dt}\:=\mathrm{2}\left(\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}} \:\:\left(\frac{\mathrm{1}}{{t}−\mathrm{1}}−\frac{\mathrm{1}}{{t}+\mathrm{1}}\right){dt} \\ $$$$=\mathrm{2}\left(\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\left[{ln}\mid\frac{{t}−\mathrm{1}}{{t}+\mathrm{1}}\mid\right]_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} ^{\sqrt{\frac{\mathrm{2}}{\mathrm{5}}}} \:?=….{so}\:{A}\:{is}\:{known} \\ $$$$ \\ $$