Question Number 120599 by MagdiRagheb last updated on 01/Nov/20
![Let x = u^6 dx = 6u^5 du I = ∫(u^3 /((1+u^2 )^2 )) ×6u^5 du =6 ∫(u^8 /(1+2u^2 +u^4 )) du =6 ∫[((−4)/(u^2 +1))+(1/((1+u^2 )^2 ))+u^4 −2u^2 +3]du =6[−4tan^(−1) (u) + I_1 + (u^5 /5) − (2/3)u^3 +3u]+c I_1 = ∫(1/((1+u^2 )^2 )) du Put u = tan z du = sec^2 z dz I_1 = ∫(1/(sec^4 z))×sec^2 z dz = ∫cos^2 z dz = ∫(1/2)(cos 2z + 1)dz = (1/2)((1/2) sin 2z + z) = (1/2)×(u/(1+u^2 )) + (1/2) tan^(−1) u I = −4 tan^(−1) u + (u/(2(1+u^2 ))) + (1/2) tan^(−1) u +(u^5 /5) − (2/3)u^3 + 3u + c = −(7/2) tan^(−1) u + (u/(2(1+u^2 ))) + (1/5)u^5 + 3u + c = −(7/2) tan^(−1) ( ^6 (√x)) + ((x)^(1/6) /(2(1+(x^2 )^(1/6) ))) + (1/5) (x^5 )^(1/6) + 3 (x)^(1/6) + c](https://www.tinkutara.com/question/Q120599.png)
$${Let}\:{x}\:=\:{u}^{\mathrm{6}} \:\:\:\:\:\:\:\:\:{dx}\:=\:\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$${I}\:=\:\int\frac{{u}^{\mathrm{3}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:×\mathrm{6}{u}^{\mathrm{5}} \:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\frac{{u}^{\mathrm{8}} }{\mathrm{1}+\mathrm{2}{u}^{\mathrm{2}} +{u}^{\mathrm{4}} }\:{du} \\ $$$$\:\:\:=\mathrm{6}\:\int\left[\frac{−\mathrm{4}}{{u}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }+{u}^{\mathrm{4}} −\mathrm{2}{u}^{\mathrm{2}} +\mathrm{3}\right]{du} \\ $$$$\:\:\:=\mathrm{6}\left[−\mathrm{4}{tan}^{−\mathrm{1}} \left({u}\right)\:+\:{I}_{\mathrm{1}} \:+\:\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}} \:+\mathrm{3}{u}\right]+{c} \\ $$$${I}_{\mathrm{1}} \:=\:\int\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{du} \\ $$$${Put}\:{u}\:=\:{tan}\:{z}\:\:\:\:\:{du}\:=\:{sec}^{\mathrm{2}} {z}\:{dz} \\ $$$${I}_{\mathrm{1}} \:=\:\int\frac{\mathrm{1}}{{sec}^{\mathrm{4}} {z}}×{sec}^{\mathrm{2}} {z}\:{dz}\:=\:\int{cos}^{\mathrm{2}} {z}\:{dz} \\ $$$$\:\:\:\:\:=\:\int\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\:\mathrm{2}{z}\:+\:\mathrm{1}\right){dz} \\ $$$$\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{sin}\:\mathrm{2}{z}\:+\:{z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} {u} \\ $$$${I}\:=\:−\mathrm{4}\:{tan}^{−\mathrm{1}} {u}\:+\:\frac{{u}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} {u}\:+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}{u}^{\mathrm{3}} \:+\:\mathrm{3}{u}\:+\:{c} \\ $$$$\:\:\:=\:−\frac{\mathrm{7}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} {u}\:+\:\frac{{u}}{\mathrm{2}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{1}}{\mathrm{5}}{u}^{\mathrm{5}} \:+\:\mathrm{3}{u}\:+\:{c} \\ $$$$\:\:\:=\:−\frac{\mathrm{7}}{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left(\overset{\mathrm{6}} {\:}\sqrt{{x}}\right)\:+\:\frac{\sqrt[{\mathrm{6}}]{{x}}}{\mathrm{2}\left(\mathrm{1}+\sqrt[{\mathrm{6}}]{{x}^{\mathrm{2}} }\right)}\:+\:\frac{\mathrm{1}}{\mathrm{5}}\:\sqrt[{\mathrm{6}}]{{x}^{\mathrm{5}} }\:+\:\mathrm{3}\:\sqrt[{\mathrm{6}}]{{x}}\:+\:{c} \\ $$