Question Number 45706 by Meritguide1234 last updated on 15/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18
![trying to solve... ∫(((1+x^4 ))/((1−x^4 )(√(1+x^4 )) ))dx ∫((1+x^4 )/(x^2 ((1/x^2 )−x^2 )×x(√((1/x^2 )+x^2 )) ))dx ∫(((1/x^3 )+x)/(((1/x^2 )−x^2 )×{((1/x^2 )+x^2 )^2 }^(1/4) ))dx ∫((((1/x^3 )+x)dx)/(((1/x^2 )−x^2 )×{((1/x^2 )−x^2 )^2 +4}^(1/4) )) t=(1/x^2 )−x^2 (dt/dx)=((−2)/x^3 )−2x (dt/(−2))=((1/x^3 )+x)dx ∫(dt/(−2×t×(t^2 +4)^(1/4) )) ((−1)/2)∫((tdt)/(t^2 ×(t^2 +4)^(1/4) )) y^4 =t^2 +4 4y^3 ×(dy/dt)=2t ((−1)/2)∫((2y^3 dy)/((y^4 −4)×y)) =−1∫((y^2 dy)/((y^2 +1)(y^2 −1))) =((−1)/2)∫((y^2 +1+y^2 −1)/((y^2 +1)(y^2 −1)))dy ((−1)/2)[∫(dy/(y^2 −1))+∫(dy/(y^2 +1))] =((−1)/2)[{(1/2)∫(((y+1)−(y−1})/((y+1)(y−1)))+∫(dy/(y^2 +1))] ((−1)/2)[{(1/2)×∫(dy/(y−1))−(1/2)∫(dy/(y+1))+∫(dy/(y^2 +1)) dy] =((−1)/2)[(1/2)ln(((y−1)/(y+1)))+tan^(−1) y] =((−1)/2)[(1/2)ln{(((t^2 +4)^(1/4) −1)/((t^2 +4)^(1/4) +1))}+tan^(−1) (t^2 +4)^(1/4) ]+c =((−1)/2)[(1/2)ln∣(({((1/x^2 )−x^2 )^2 +4}^(1/4) −1)/({((1/x^2 )−x^2 )^2 +4}^(1/4) +2))∣+tan^(−1) {((1/x^2 )−x)^2 +4}^(1/4) +c](https://www.tinkutara.com/question/Q45881.png)
$${trying}\:{to}\:{solve}… \\ $$$$\int\frac{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)}{\left(\mathrm{1}−{x}^{\mathrm{4}} \right)\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }\:}{dx} \\ $$$$\int\frac{\mathrm{1}+{x}^{\mathrm{4}} }{{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \right)×{x}\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} }\:}{dx} \\ $$$$\int\frac{\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}}{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \right)×\left\{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \right\}^{\frac{\mathrm{1}}{\mathrm{4}}} }{dx} \\ $$$$\int\frac{\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}\right){dx}}{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \right)×\left\{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$${t}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \:\:\:\:\frac{{dt}}{{dx}}=\frac{−\mathrm{2}}{{x}^{\mathrm{3}} }−\mathrm{2}{x}\:\:\:\:\frac{{dt}}{−\mathrm{2}}=\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}\right){dx} \\ $$$$\int\frac{{dt}}{−\mathrm{2}×{t}×\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{tdt}}{{t}^{\mathrm{2}} ×\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$${y}^{\mathrm{4}} ={t}^{\mathrm{2}} +\mathrm{4}\:\:\:\:\mathrm{4}{y}^{\mathrm{3}} ×\frac{{dy}}{{dt}}=\mathrm{2}{t} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{y}^{\mathrm{3}} {dy}}{\left({y}^{\mathrm{4}} −\mathrm{4}\right)×{y}} \\ $$$$=−\mathrm{1}\int\frac{{y}^{\mathrm{2}} {dy}}{\left({y}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{y}^{\mathrm{2}} +\mathrm{1}+{y}^{\mathrm{2}} −\mathrm{1}}{\left({y}^{\mathrm{2}} +\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{1}\right)}{dy} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\left[\int\frac{{dy}}{{y}^{\mathrm{2}} −\mathrm{1}}+\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left[\left\{\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({y}+\mathrm{1}\right)−\left({y}−\mathrm{1}\right\}}{\left({y}+\mathrm{1}\right)\left({y}−\mathrm{1}\right)}+\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}}\right]\right. \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\left[\left\{\frac{\mathrm{1}}{\mathrm{2}}×\int\frac{{dy}}{{y}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dy}}{{y}+\mathrm{1}}+\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{1}}\:{dy}\right]\right. \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{y}−\mathrm{1}}{{y}+\mathrm{1}}\right)+{tan}^{−\mathrm{1}} {y}\right] \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\frac{\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}}\right\}+{tan}^{−\mathrm{1}} \left({t}^{\mathrm{2}} +\mathrm{4}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \right]+{c} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\frac{\left\{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} −\mathrm{1}}{\left\{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{2}}\mid+{tan}^{−\mathrm{1}} \left\{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−{x}\right)^{\mathrm{2}} +\mathrm{4}\right\}^{\frac{\mathrm{1}}{\mathrm{4}}} +{c}\right. \\ $$$$ \\ $$
Commented by Meritguide1234 last updated on 18/Oct/18
$${nice}\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 18/Oct/18
$${thank}\:{you}\:{sir}… \\ $$