Question Number 202388 by Calculusboy last updated on 25/Dec/23
$$\:\:\boldsymbol{{P}}\:\boldsymbol{{rove}}\:\boldsymbol{{that}}:\:\:\:\:\int\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{b}}^{\mathrm{4}} +\mathrm{2}\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{c}}}=\frac{\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}\right)}{\:\sqrt{\mathrm{2}}\sqrt{\boldsymbol{{a}}}\sqrt{\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} }}+\boldsymbol{{C}} \\ $$$$\boldsymbol{{if}}\:\:\boldsymbol{{a}}\centerdot\left(\boldsymbol{{c}}+\boldsymbol{{b}}^{\mathrm{4}} \right)>\mathrm{0} \\ $$$$ \\ $$
Answered by witcher3 last updated on 26/Dec/23
$$\int\frac{\mathrm{dx}}{\mathrm{b}^{\mathrm{4}} +\mathrm{c}+\mathrm{2ax}^{\mathrm{2}} }=\int\frac{\mathrm{dx}}{\mathrm{2a}\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{b}^{\mathrm{4}} +\mathrm{c}}{\mathrm{2a}}\right)};\beta^{\mathrm{2}} =\frac{\mathrm{b}^{\mathrm{4}} +\mathrm{c}}{\mathrm{2a}}>\mathrm{0} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2a}}\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\beta^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2a}\beta}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\beta}\right)+\mathrm{k};\mathrm{k}\in\mathbb{R} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2a}\sqrt{\frac{\mathrm{b}^{\mathrm{4}} +\mathrm{c}}{\mathrm{2a}}}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\sqrt{\frac{\mathrm{2a}}{\mathrm{b}^{\mathrm{4}} +\mathrm{c}}}\right)+\mathrm{k} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}.\sqrt{\mathrm{a}}.\sqrt{\mathrm{b}^{\mathrm{4}} +\mathrm{c}}}.\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}\sqrt{\mathrm{2}}.\sqrt{\mathrm{a}}}{\:\sqrt{\mathrm{c}+\mathrm{b}^{\mathrm{4}} }}\right)+\mathrm{k} \\ $$
Commented by Calculusboy last updated on 26/Dec/23
$$\boldsymbol{{nice}}\:\boldsymbol{{solution}} \\ $$