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Question Number 168978 by MikeH last updated on 22/Apr/22

Evaluate (a) ∫((t−2)/(t−3(√(2t−4)) +2)) dt (b) ∫((3z)/((1−4z−2z^2 )^2 )) dz

$$\mathrm{Evaluate}\: \\ $$$$\left(\mathrm{a}\right)\:\int\frac{{t}−\mathrm{2}}{{t}−\mathrm{3}\sqrt{\mathrm{2}{t}−\mathrm{4}}\:+\mathrm{2}}\:{dt}\: \\ $$$$\left(\mathrm{b}\right)\:\int\frac{\mathrm{3}{z}}{\left(\mathrm{1}−\mathrm{4}{z}−\mathrm{2}{z}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dz} \\ $$

Answered by floor(10²Eta[1]) last updated on 22/Apr/22

(a): (√(2t−4))=u⇒dt=udu, t=(u^2 /2)+2 ∫((u^3 /2)/((u^2 /2)+4−3u))du=∫(u^3 /(u^2 −6u+8))du =∫(u+6)du+∫((28u−48)/(u^2 −6u+8)) (u^2 /2)+6u+∫((28u−48)/((u−2)(u−4)))du (u^2 /2)+6u−4∫(du/(u−2))+32∫(du/(u−4)) (u^2 /2)+6u−4ln∣u−2∣+32ln∣u−4∣+C t−2+6(√(2t−4))−4ln∣(√(2t−4))−2∣+32ln∣(√(2t−4))−4∣+C

$$\left(\mathrm{a}\right): \\ $$$$\sqrt{\mathrm{2t}−\mathrm{4}}=\mathrm{u}\Rightarrow\mathrm{dt}=\mathrm{udu},\:\mathrm{t}=\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2} \\ $$$$\int\frac{\frac{\mathrm{u}^{\mathrm{3}} }{\mathrm{2}}}{\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{4}−\mathrm{3u}}\mathrm{du}=\int\frac{\mathrm{u}^{\mathrm{3}} }{\mathrm{u}^{\mathrm{2}} −\mathrm{6u}+\mathrm{8}}\mathrm{du} \\ $$$$=\int\left(\mathrm{u}+\mathrm{6}\right)\mathrm{du}+\int\frac{\mathrm{28u}−\mathrm{48}}{\mathrm{u}^{\mathrm{2}} −\mathrm{6u}+\mathrm{8}} \\ $$$$\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6u}+\int\frac{\mathrm{28u}−\mathrm{48}}{\left(\mathrm{u}−\mathrm{2}\right)\left(\mathrm{u}−\mathrm{4}\right)}\mathrm{du} \\ $$$$\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6u}−\mathrm{4}\int\frac{\mathrm{du}}{\mathrm{u}−\mathrm{2}}+\mathrm{32}\int\frac{\mathrm{du}}{\mathrm{u}−\mathrm{4}} \\ $$$$\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{6u}−\mathrm{4ln}\mid\mathrm{u}−\mathrm{2}\mid+\mathrm{32ln}\mid\mathrm{u}−\mathrm{4}\mid+\mathrm{C} \\ $$$$\mathrm{t}−\mathrm{2}+\mathrm{6}\sqrt{\mathrm{2t}−\mathrm{4}}−\mathrm{4ln}\mid\sqrt{\mathrm{2t}−\mathrm{4}}−\mathrm{2}\mid+\mathrm{32ln}\mid\sqrt{\mathrm{2t}−\mathrm{4}}−\mathrm{4}\mid+\mathrm{C} \\ $$

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