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

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

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

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

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