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Question Number 133972 by liberty last updated on 26/Feb/21

$$\mathcal{H}=\int \frac{{(2\mathrm{x}-1)}^7}{{(2\mathrm{x}+1)}^9}\mathrm{dx}$$

Answered by EDWIN88 last updated on 26/Feb/21

$$\mathcal{H}=\int {\left(\frac{2\mathrm{x}-1}{2\mathrm{x}+1}\right)}^7.\frac{\mathrm{dx}}{{(2\mathrm{x}+1)}^2}$$$$\mathrm{change}\mathrm{of}\mathrm{variable}\mathrm{let}\mathrm{u}=\frac{2\mathrm{x}-1}{2\mathrm{x}+1}$$$$\mathrm{du}=\frac{4}{{(2\mathrm{x}+1)}^2}\mathrm{dx};\frac{\mathrm{dx}}{{(2\mathrm{x}+1)}^2}=\frac{\mathrm{du}}{4}$$$$\mathcal{H}=\int \frac{{\mathrm{u}}^7}{4}\mathrm{du}=\frac{{\mathrm{u}}^8}{32}+\mathrm{c}=\underset{―}{\frac{1}{32}{\left(\frac{2\mathrm{x}-1}{2\mathrm{x}+1}\right)}^8+\mathrm{c}}$$$$$$

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