Question Number 209543 by Spillover last updated on 13/Jul/24
$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{4}} \\ $$$$ \\ $$
Commented by Frix last updated on 14/Jul/24
$$\mathrm{Done}\:\mathrm{several}\:\mathrm{times}\:\mathrm{before} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{tan}^{−\mathrm{1}} \:\left({x}−\mathrm{1}\right)\:+\mathrm{tan}^{−\mathrm{1}} \:\left({x}+\mathrm{1}\right)\right)+{C} \\ $$
Commented by Spillover last updated on 14/Jul/24
$${correct} \\ $$
Answered by Spillover last updated on 14/Jul/24
![∫(dx/(x^4 +4))=∫(dx/((x^2 −2x+2)(x^2 +2x+2))) (1/((x^2 −2x+2)(x^2 +2x+2)))=((Ax+B)/((x^2 −2x+2)))+((Cx+D)/((x^2 +2x+2))) =((Ax+B)/((x^2 −2x+2)))+((Cx+D)/((x^2 +2x+2))) (Ax+B)(x^2 +2x+2)+(Cx+D)(x^2 −2x+2)=1 (A+C)x^3 +(2A+B−2C+D)x^2 +(2A+2B+2C−2D)x+2B+2D A+C=1 2A+B−2C+D=0 2A+2B+2C−2D)=0 2B+2D=0 A=−(1/8) B=(1/4) C=(1/8) D=(1/4) =((Ax+B)/((x^2 −2x+2)))+((Cx+D)/((x^2 +2x+2))) =((−(1/8)x+(1/4))/((x^2 −2x+2)))+(((1/8)x+(1/4))/((x^2 +2x+2))) ∫[((−(1/8)x+(1/4))/((x^2 −2x+2)))+(((1/8)x+(1/4))/((x^2 +2x+2)))]dx (1/8)∫((−x+2)/((x^2 −2x+2)))dx+(1/8)∫((x+2)/((x^2 +2x+2)))dx I_1 =(1/8)∫((−x+2)/((x^2 −2x+2)))dx I_1 =(1/8)∫((−(1/2)(2x−2)+1)/((x^2 −2x+2)))dx=(1/8)[∫((−(1/2)(2x−2))/((x^2 −2x+2)))+∫(dx/((x^2 −2x+2)))dx= (1/8)[∫((−(1/2)(2x−2))/((x^2 −2x+2)))+∫(dx/((x^2 −2x+2)))dx I_1 =−(1/(16))ln (x^2 −2x+2)+(1/8)tan^(−1) (x−1) I_2 =(1/8)∫((x+2)/((x^2 +2x+2)))dx I_2 =(1/8)∫(((1/2)(2x+2)+1)/((x^2 +2x+2)))dx=(1/8)[∫(((1/2)(2x+2))/((x^2 +2x+2)))+∫(dx/((x^2 +2x+2)))]dx (1/8)[∫(((1/2)(2x+2))/((x^2 +2x+2)))+∫(dx/((x^2 +2x+2)))]dx I_2 =−(1/(16))ln (x^2 +2x+2)+(1/8)tan^(−1) (x+1) I_1 =−(1/(16))ln (x^2 −2x+2)+(1/8)tan^(−1) (x−1) I_2 =(1/(16))ln (x^2 +2x+2)+(1/8)tan^(−1) (x+1) I=I_1 +I_2 =(1/(16))ln ((x^2 +2x+2)/(x^2 −2x+2)) +(1/8)[tan^(−1) (x−1) +tan^(−1) (x+1)]+A](https://www.tinkutara.com/question/Q209554.png)
$$\int\frac{{dx}}{{x}^{\mathrm{4}} +\mathrm{4}}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}=\frac{{Ax}+{B}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{{Cx}+{D}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$=\frac{{Ax}+{B}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{{Cx}+{D}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\left({Ax}+{B}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)+\left({Cx}+{D}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)=\mathrm{1} \\ $$$$\left({A}+{C}\right){x}^{\mathrm{3}} +\left(\mathrm{2}{A}+{B}−\mathrm{2}{C}+{D}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{A}+\mathrm{2}{B}+\mathrm{2}{C}−\mathrm{2}{D}\right){x}+\mathrm{2}{B}+\mathrm{2}{D} \\ $$$${A}+{C}=\mathrm{1} \\ $$$$\mathrm{2}{A}+{B}−\mathrm{2}{C}+{D}=\mathrm{0} \\ $$$$\left.\mathrm{2}{A}+\mathrm{2}{B}+\mathrm{2}{C}−\mathrm{2}{D}\right)=\mathrm{0} \\ $$$$\mathrm{2}{B}+\mathrm{2}{D}=\mathrm{0} \\ $$$${A}=−\frac{\mathrm{1}}{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{B}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{C}=\frac{\mathrm{1}}{\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:{D}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$=\frac{{Ax}+{B}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{{Cx}+{D}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$=\frac{−\frac{\mathrm{1}}{\mathrm{8}}{x}+\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\frac{\mathrm{1}}{\mathrm{8}}{x}+\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)} \\ $$$$\int\left[\frac{−\frac{\mathrm{1}}{\mathrm{8}}{x}+\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\frac{\mathrm{1}}{\mathrm{8}}{x}+\frac{\mathrm{1}}{\mathrm{4}}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}\right]{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\int\frac{−{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx}+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{dx} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{8}}\int\frac{−{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{8}}\int\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}−\mathrm{2}\right)+\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{8}}\left[\int\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}−\mathrm{2}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx}=\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\left[\int\frac{−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}−\mathrm{2}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}+\int\frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)}{dx}\right. \\ $$$${I}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right) \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{x}+\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{dx} \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{2}\right)+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{8}}\left[\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{2}\right)}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}\right]{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\left[\int\frac{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\mathrm{2}\right)}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}\right]{dx} \\ $$$${I}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right) \\ $$$${I}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right) \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{−\mathrm{1}} \left({x}+\mathrm{1}\right) \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{tan}^{−\mathrm{1}} \:\left({x}−\mathrm{1}\right)\:+\mathrm{tan}^{−\mathrm{1}} \:\left({x}+\mathrm{1}\right)\right]+{A} \\ $$