Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 26486 by A1B1C1D1 last updated on 26/Dec/17

$${\int }_{-1}^1\frac{\mathrm{xdx}}{{({\mathrm{x}}^2+1)}^2}$$

Commented by kaivan.ahmadi last updated on 26/Dec/17

$$\mathrm{u}={\mathrm{x}}^2+1\Rightarrow \mathrm{du}=2\mathrm{xdx}$$$$\Rightarrow \frac{1}{2}\int \frac{\mathrm{du}}{{\mathrm{u}}^2}=1/2(-\frac{1}{\mathrm{u}})=-1/2\left(\frac{1}{{\mathrm{x}}^2+1}\right){\mid }_{-1}^1=$$$$-1/2(\frac{1}{2}-\frac{1}{2})=0$$

Commented by A1B1C1D1 last updated on 26/Dec/17

$$\mathrm{Thanks}.$$

Commented by mrW1 last updated on 26/Dec/17

$$foranyf\left(x\right):$$$$iff(-x)=-f\left(x\right)$$$${\int }_{-a}^{a}f\left(x\right)dx={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$$$$=-{\int }_0^{a}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx$$$$=0$$

Commented by A1B1C1D1 last updated on 26/Dec/17

$$\mathrm{Thanks}$$

Commented by abdo imad last updated on 26/Dec/17

$$ingeneraliffisintegrableandodd(f(-x)=-f\left(x\right))$$$$on[-a,a]\Rightarrow {\int }_{-a}^{a}f\left(x\right)dx=0$$

Answered by jota@ last updated on 26/Dec/17

$$=0.Thefunctionisimpar.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com