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Question Number 11406 by tawa last updated on 24/Mar/17

$$\mathrm{Evaluate}:{\int }_1^3\frac{\mathrm{x}-1}{{(\mathrm{x}+1)}^2}\mathrm{dx}$$

Answered by sm3l2996 last updated on 24/Mar/17

$$\frac{\mathrm{x}-1}{{(\mathrm{x}+1)}^2}=\frac{\mathrm{a}}{\mathrm{x}+1}+\frac{\mathrm{b}}{{(\mathrm{x}+1)}^2}$$$$\mathrm{b}=-2;\mathrm{a}=1$$$$\mathrm{I}={\int }_1^3\frac{\mathrm{x}-1}{{(\mathrm{x}+1)}^2}\mathrm{dx}={\int }_1^3\frac{\mathrm{dx}}{\mathrm{x}+1}-2{\int }_1^3\frac{\mathrm{dx}}{{(\mathrm{x}+1)}^2}$$$$={[\ln \mid \mathrm{x}+1\mid ]}_1^3+2{\left[\frac{1}{\mathrm{x}+1}\right]}_1^3$$$$=\ln \left(2\right)-\frac{1}{2}$$

Commented by tawa last updated on 24/Mar/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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