Question Number 86288 by john santu last updated on 28/Mar/20
$$\int\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\mathrm{1}/\mathrm{4}} } \\ $$
Answered by TANMAY PANACEA. last updated on 28/Mar/20
$${x}^{\mathrm{2}} ={tana} \\ $$$$\mathrm{2}{xdx}={sec}^{\mathrm{2}} {ada} \\ $$$$\int\frac{{sec}^{\mathrm{2}} {ada}}{\mathrm{2}\sqrt{{tana}}\:×\left({sec}^{\mathrm{2}} {a}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}×\frac{\sqrt{{sina}}}{\:\sqrt{{cosa}}}×\frac{\mathrm{1}}{\:\sqrt{{cosa}}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{cosa}}{\left(\mathrm{1}−{sin}^{\mathrm{2}} {a}\right)\sqrt{{sina}}}{da} \\ $$$${b}^{\mathrm{2}} ={sina} \\ $$$$\mathrm{2}{bdb}={cosada} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{bdb}}{\left(\mathrm{1}−{b}^{\mathrm{2}} \right){b}} \\ $$$$\int\frac{{db}}{\mathrm{1}−{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{b}}{\mathrm{1}−{b}}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+\sqrt{{sina}}}{\mathrm{1}−\sqrt{{sina}}}\right)+{c} \\ $$$${tana}={x}^{\mathrm{2}} \rightarrow{sina}=\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{ans}}\:=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{ln}}\left(\frac{\mathrm{1}+\frac{\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }}{\mathrm{1}−\frac{\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} }}\right)+\boldsymbol{{c}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{ln}}\left(\frac{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} +\boldsymbol{{x}}}{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} −\boldsymbol{{x}}}\right)+\boldsymbol{{c}} \\ $$$$ \\ $$