Question Number 26486 by A1B1C1D1 last updated on 26/Dec/17
$$\int_{\:−\mathrm{1}} ^{\:\:\mathrm{1}} \frac{\mathrm{xdx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Commented by kaivan.ahmadi last updated on 26/Dec/17
$$\mathrm{u}=\mathrm{x}^{\mathrm{2}} +\mathrm{1}\Rightarrow\mathrm{du}=\mathrm{2xdx} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} }=\mathrm{1}/\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{u}}\right)=−\mathrm{1}/\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\right)\mid_{−\mathrm{1}} ^{\mathrm{1}} = \\ $$$$−\mathrm{1}/\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0} \\ $$
Commented by A1B1C1D1 last updated on 26/Dec/17
$$\mathrm{Thanks}. \\ $$
Commented by mrW1 last updated on 26/Dec/17
$${for}\:{any}\:{f}\left({x}\right): \\ $$$${if}\:{f}\left(−{x}\right)=−{f}\left({x}\right) \\ $$$$\int_{−{a}} ^{\:\:{a}} {f}\left({x}\right){dx}=\int_{−{a}} ^{\:\mathrm{0}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\:{a}} {f}\left({x}\right){dx} \\ $$$$=−\int_{\mathrm{0}} ^{\:{a}} {f}\left({x}\right){dx}+\int_{\mathrm{0}} ^{\:{a}} {f}\left({x}\right){dx} \\ $$$$=\mathrm{0} \\ $$
Commented by A1B1C1D1 last updated on 26/Dec/17
$$\mathrm{Thanks} \\ $$
Commented by abdo imad last updated on 26/Dec/17
![in general if f is integrable and odd(f(−x)=−f(x)) on [−a,a]⇒∫_(−a) ^a f(x)dx=0](https://www.tinkutara.com/question/Q26546.png)
$${in}\:{general}\:{if}\:{f}\:{is}\:{integrable}\:{and}\:{odd}\left({f}\left(−{x}\right)=−{f}\left({x}\right)\right) \\ $$$${on}\:\left[−{a},{a}\right]\Rightarrow\int_{−{a}} ^{{a}} {f}\left({x}\right){dx}=\mathrm{0} \\ $$
Answered by jota@ last updated on 26/Dec/17
$$=\mathrm{0}.\:{The}\:\:{function}\:{is}\:{impar}. \\ $$