Question Number 211578 by MrGaster last updated on 13/Sep/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \frac{\boldsymbol{{x}}}{\:\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} }}\boldsymbol{{dx}}. \\ $$$$ \\ $$
Answered by lepuissantcedricjunior last updated on 13/Sep/24
![∫_0 ^∞ ((xdx)/( (√(1+x^4 ))))=k k=∫_0 ^∞ ((d(x^2 ))/(2(√(1+(x^2 )^2 )))) ^(x^2 =tant=>d(x^2 )=(1+tan^2 t)dt) k=∫_0 ^(𝛑/2) ((1+tan^2 t)/(1/(cost)))=∫_0 ^(𝛑/2) (cost+((sin^2 )/(cost)))dt k=1+∫_0 ^(𝛑/2) ((1/(cost))−cost)dt k=∫_0 ^(𝛑/2) ((1/2)(((cost)/(1−sint))+((cost)/(1+sint))))dt k=(1/2)[ln(((1+sint)/(1−sint)))]_0 ^(𝛑/2) =−∞](https://www.tinkutara.com/question/Q211592.png)
$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{xdx}}}{\:\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} }}=\boldsymbol{{k}} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{{d}}\left(\boldsymbol{{x}}^{\mathrm{2}} \right)}{\mathrm{2}\sqrt{\mathrm{1}+\left(\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{2}} }}\:\:\overset{\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{tant}}=>\boldsymbol{{d}}\left(\boldsymbol{{x}}^{\mathrm{2}} \right)=\left(\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{t}}\right)\boldsymbol{{dt}}} {\:} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\boldsymbol{\pi}/\mathrm{2}} \frac{\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{t}}}{\frac{\mathrm{1}}{\boldsymbol{{cost}}}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\boldsymbol{{cost}}+\frac{\boldsymbol{{sin}}^{\mathrm{2}} }{\boldsymbol{{cost}}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}=\mathrm{1}+\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\boldsymbol{{cost}}}−\boldsymbol{{cost}}\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\boldsymbol{{cost}}}{\mathrm{1}−\boldsymbol{{sint}}}+\frac{\boldsymbol{{cost}}}{\mathrm{1}+\boldsymbol{{sint}}}\right)\right)\boldsymbol{{dt}} \\ $$$$\boldsymbol{{k}}=\frac{\mathrm{1}}{\mathrm{2}}\left[\boldsymbol{{ln}}\left(\frac{\mathrm{1}+\boldsymbol{{sint}}}{\mathrm{1}−\boldsymbol{{sint}}}\right)\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} =−\infty \\ $$
Answered by Frix last updated on 13/Sep/24
![∫(x/( (√(1+x^4 ))))dx =^([t=x^2 +(√(x^4 +1))]) (1/2)∫(dt/t)=((ln t)/2)= =((ln (x^2 +(√(x^4 +1))))/2)+C ∫_0 ^∞ (x/( (√(1+x^4 ))))dx diverges](https://www.tinkutara.com/question/Q211585.png)
$$\int\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}{dx}\:\overset{\left[{t}={x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right]} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}}=\frac{\mathrm{ln}\:{t}}{\mathrm{2}}= \\ $$$$=\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}\right)}{\mathrm{2}}+{C} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}{dx}\:\mathrm{diverges} \\ $$