Question Number 57421 by Abdo msup. last updated on 03/Apr/19
$${calculate}\:\:\int_{−\mathrm{1}} ^{\mathrm{1}} \:\:\:\frac{\left({x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:+{e}^{{x}} }{{e}^{{x}} \:+\mathrm{1}}{dx} \\ $$
Answered by einsteindrmaths@hotmail.fr last updated on 04/Apr/19
![=∫_(−1) ^1 (((x^4 +x^2 +1)^2 +e^x )/(e^x +1))dx=∫_(−1) ^1 (((x^4 +x^2 +1)^2 )/(e^x +1))dx+∫_(−1) ^1 (e^x /(e^x +1))dx the seconde integral is easy to evaluat ∫(e^x /(e^x +1))dx=ln(e^x +1)+c lets find ∫_(−1) ^1 (((x^4 +x^2 +1)^2 )/(e^x +1))dx=∫_(−1) ^0 (((x^4 +x^2 +1)^2 )/(e^x +1))dx+∫_0 ^1 (((x^4 +x^2 +1)^4 )/(e^x +1))dx we put y=−x in the firste we get ∫_0 ^1 (((x^4 +x^2 +1)^2 )/(e^(−y) +1))dy+∫_0 ^1 (((x^4 +x^2 +1)^2 )/(e^x +1))dx =∫_0 ^1 [ (((x^4 +x^2 +1)^2 )/(e^x +1))+(((x^4 +x^2 +1)^2 )/(e^(−x) +1)) ] dx=∫_0 ^1 (x^4 +x^2 +1)^2 dx=∫_0 ^1 (x^8 +2x^6 +3x^4 +2x^2 +1)dx easy to evaluat](Q57436.png)
$$=\underset{−\mathrm{1}} {\int}\overset{\mathrm{1}} {\:}\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +{e}^{{x}} }{{e}^{{x}} +\mathrm{1}}{dx}=\underset{−\mathrm{1}} {\int}\overset{\mathrm{1}} {\:}\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{{x}} +\mathrm{1}}{dx}+\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{{e}^{{x}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$${the}\:{seconde}\:{integral}\:{is}\:{easy}\:{to}\:{evaluat}\: \\ $$$$\int\frac{{e}^{{x}} }{{e}^{{x}} +\mathrm{1}}{dx}={ln}\left({e}^{{x}} +\mathrm{1}\right)+{c} \\ $$$$ \\ $$$${lets}\:{find}\:\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{{x}} +\mathrm{1}}{dx}=\underset{−\mathrm{1}} {\overset{\mathrm{0}} {\int}}\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{{x}} +\mathrm{1}}{dx}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$${we}\:{put}\:{y}=−{x}\:{in}\:{the}\:{firste}\:{we}\:{get}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{−{y}} +\mathrm{1}}{dy}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{{x}} +\mathrm{1}}{dx} \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left[\:\:\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{{x}} +\mathrm{1}}+\frac{\left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{−{x}} +\mathrm{1}}\:\right]\:\:{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} {dx}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left({x}^{\mathrm{8}} +\mathrm{2}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){dx} \\ $$$${easy}\:{to}\:{evaluat}\: \\ $$