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x-1-x-2-1-dx-




Question Number 180935 by SANOGO last updated on 19/Nov/22
∫_ ((x+1)/(x^2 +1))dx
$$\int_{} \frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$
Answered by Frix last updated on 19/Nov/22
∫((x+1)/(x^2 +1))dx=(1/2)∫((2x)/(x^2 +1))dx+∫(dx/(x^2 +1))=  =((ln (x^2 +1))/2)+tan^(−1)  x +C
$$\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}+\mathrm{tan}^{−\mathrm{1}} \:{x}\:+{C} \\ $$
Commented by SANOGO last updated on 19/Nov/22
merci bien
$${merci}\:{bien} \\ $$
Answered by Socracious last updated on 19/Nov/22
$$ \\ $$

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