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Question Number 83822 by Power last updated on 06/Mar/20

Commented by mathmax by abdo last updated on 07/Mar/20

$$I=\int \frac{dx}{\sqrt{2x-1}{-}^4\sqrt{2x-1}}cha7gement{}^4\sqrt{2x-1}=tgive$$$$2x-1=t^4\Rightarrow 2x=1+t^4\Rightarrow 2dx=4t^{3}dt\Rightarrow dx=2t^{3}dt\Rightarrow$$$$I=\int \frac{2t^{3}dt}{\sqrt{1+t^4-1}-t}=2\int \frac{t^3}{t^2-t}dt=2\int \frac{t^2}{t-1}dt$$$$=2\int \frac{t^2-1+1}{t-1}dt=2\int (t+1)dt+2\int \frac{dt}{t-1}$$$$=t^2+2t+2ln\mid t-1\mid +C$$$$=\sqrt{2x-1}+2{}^4\sqrt{2x-1}+2ln{\mid }^4\sqrt{2x-1}-1\mid +C$$

Answered by john santu last updated on 06/Mar/20

$$\mathrm{let}\sqrt[4]{2\mathrm{x}-1}=\mathrm{u}\Rightarrow {\mathrm{u}}^4=2\mathrm{x}-1$$$${4\mathrm{u}}^3\mathrm{du}=2\mathrm{dx}$$$$\int \frac{{2\mathrm{u}}^3\mathrm{du}}{{\mathrm{u}}^2-\mathrm{u}}=\int \frac{{2\mathrm{u}}^2}{\mathrm{u}-1}\mathrm{du}=$$$$\frac{2}{3}{\mathrm{u}}^2+2\mathrm{u}+\int \frac{2\mathrm{du}}{\mathrm{u}-1}=$$$$\frac{2}{3}\sqrt{2\mathrm{x}-1}+2\sqrt[4]{2\mathrm{x}-1}+2\ln \mid \sqrt[4]{2\mathrm{x}-1}-1\mid +\mathrm{c}$$

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