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Question Number 92055 by 675480065 last updated on 04/May/20

$$\int \left(\frac{\mathrm{x}+1}{{({\mathrm{x}}^2+4\mathrm{x}+5)}^2}\right)\mathrm{dx}$$

Commented by Prithwish Sen 1 last updated on 04/May/20

$$\int \frac{\mathrm{x}+1}{{({(\mathrm{x}+2)}^2+1)}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{x}-2$$$$\int \frac{\mathrm{x}-1}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=\frac{1}{2}\int \frac{2\mathrm{x}}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}-\int \frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^2}$$$$=\frac{-1}{2({\mathrm{x}}^2+1)}-\frac{\mathrm{x}}{2(1+{\mathrm{x}}^2)}-\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{c}$$$$=-\frac{(\mathrm{x}+3)}{2({\mathrm{x}}^2+4\mathrm{x}+5)}-\frac{1}{2}{\tan }^{-1}(\mathrm{x}+2)+\mathrm{c}$$$$\mathrm{I}\mathrm{just}\mathrm{forgot}\mathrm{to}\mathrm{do}\mathrm{the}\mathrm{inverse}\mathrm{transformation}.$$$$\mathrm{thank}\mathrm{you}.\mathbf{please}\mathbf{check}.$$

Commented by 675480065 last updated on 04/May/20

$$\mathrm{i}\mathrm{dont}\mathrm{get}\mathrm{u}$$

Commented by Prithwish Sen 1 last updated on 04/May/20

$$\frac{\mathrm{x}+1}{{({\mathrm{x}}^2+4\mathrm{x}+5)}^2}\mathrm{and}\frac{\mathrm{x}-1}{{({\mathrm{x}}^2+1)}^2}\mathrm{are}\mathrm{the}\mathrm{same}\mathrm{function}$$$$\mathrm{only}\mathrm{the}\mathrm{function}\mathrm{has}\mathrm{been}\mathrm{a}\mathrm{horizontal}\mathrm{shifting}$$$$\mathrm{along}\mathrm{x}\mathrm{axis}\mathrm{in}\mathrm{the}\mathrm{rightward}\mathrm{direction}.\mathrm{There}$$$$\mathrm{should}\mathrm{not}\mathrm{be}\mathrm{any}\mathrm{impact}\mathrm{on}\mathrm{overall}\mathrm{calculation}.$$$$$$

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