Question Number 36633 by abdo.msup.com last updated on 03/Jun/18
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right){dx} \\ $$
Commented by abdo.msup.com last updated on 05/Jun/18
![let integrate by parts I = [ x arctan(x^2 +x+1)]_0 ^1 −∫_0 ^1 x ((2x+1)/(1 +(x^2 +x+1)^2 )) =arctan(3) −∫_0 ^1 ((2x^2 +x)/(1+x^4 +x^2 +1 +2x^3 +2x^2 +2x))dx =arctan(3) −∫_0 ^1 ((2x^2 +x)/(x^4 +2x^3 +3x^2 +2x+2))dx let decompose F(x)= ((2x^2 +x)/(x^4 +2x^3 +3x^2 +2x +2)) the roots of D_(r ) are i ,−i,−1+i,−1−i F(x)= ((2x^2 +x)/((x−i)(x+i)(x−(−1+i))( x+1+i))) =((2x^2 +x)/((x^2 +1)((x+1)^2 +1))) = ((2x^2 +x)/((x^2 +1)(x^2 +2x +2))) =((ax+b)/(x^2 +1)) +((cx+d)/(x^2 +2x +2)) lim_(x→+∞) xF(x)=0=a+c ⇒c=−a so F(x) =((ax +b)/(x^2 +1)) +((−ax +d)/(x^2 +2x +2)) F(0)=0 =b +(d/2) ⇒2b +d=0⇒d=−2b F(x) = ((ax+b)/(x^2 +1)) −((ax +2b)/(x^2 +2x+2)) F(1)= (3/(2.5)) = ((a+b)/2) −((a+2b)/5) ⇒ 3 =5a+5b −2a −4b⇒3a +b=3 F(−1) =(1/2) =((−a+b)/2) −((−a+2b)/1) ⇒ −a+b +2a −4b =1 ⇒a−3b =1⇒ a=3b +1 ⇒3(3b+1) +b =3 ⇒ 9b +3 +b =3 ⇒b=0 ⇒a =1⇒ F(x) = (x/(x^2 +1)) −(x/(x^2 +2x+2)) ∫_0 ^1 F(x)dx = ∫_0 ^1 ((xdx)/(x^2 +1)) −∫_0 ^1 ((xdx)/(x^2 +2x+2)) but ∫_0 ^1 ((xdx)/(x^2 +1)) =[(1/2)ln(x^2 +1)]_0 ^1 =((ln(2))/2) ∫_0 ^1 (x/(x^2 +2x+2))dx = (1/2) ∫_0 ^1 ((2x+2−2)/(x^2 +2x+2))dx =(1/2)[ln(x^2 +2x+2)]_0 ^1 −∫_0 ^1 (dx/(x^2 +2x+2)) =(1/2){ln(5)−ln(2)} −∫_0 ^1 (dx/(x^2 +2x +2)) ∫_0 ^1 (dx/(x^2 +2x +2)) =∫_0 ^1 (dx/((x+1)^2 +1)) =_(x+1=u) ∫_1 ^2 (du/(1+u^2 )) =arctan(2)−(π/4)⇒ ∫_0 ^1 arctan(x^2 +x+1) = arctan(3)−((ln(2))/2) + arctan(2) −(π/4) ★](https://www.tinkutara.com/question/Q36802.png)
$${let}\:{integrate}\:{by}\:{parts} \\ $$$${I}\:=\:\left[\:{x}\:{arctan}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$−\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{1}\:+\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$={arctan}\left(\mathrm{3}\right)\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}}{\mathrm{1}+{x}^{\mathrm{4}} \:+{x}^{\mathrm{2}} \:+\mathrm{1}\:+\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}{x}}{dx} \\ $$$$={arctan}\left(\mathrm{3}\right)\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}}{{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{3}} \:\:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$ \\ $$$$ \\ $$$${let}\:{decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}}{{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{3}} \:+\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}} \\ $$$${the}\:{roots}\:{of}\:{D}_{{r}\:} \:{are}\:{i}\:,−{i},−\mathrm{1}+{i},−\mathrm{1}−{i} \\ $$$${F}\left({x}\right)=\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}}{\left({x}−{i}\right)\left({x}+{i}\right)\left({x}−\left(−\mathrm{1}+{i}\right)\right)\left(\:{x}+\mathrm{1}+{i}\right)} \\ $$$$=\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} \:+{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}\right)} \\ $$$$=\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{cx}+{d}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}={a}+{c}\:\Rightarrow{c}=−{a}\:{so} \\ $$$${F}\left({x}\right)\:=\frac{{ax}\:+{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−{ax}\:+{d}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0}\:={b}\:+\frac{{d}}{\mathrm{2}}\:\Rightarrow\mathrm{2}{b}\:+{d}=\mathrm{0}\Rightarrow{d}=−\mathrm{2}{b} \\ $$$${F}\left({x}\right)\:=\:\frac{{ax}+{b}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{ax}\:+\mathrm{2}{b}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}} \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{3}}{\mathrm{2}.\mathrm{5}}\:=\:\frac{{a}+{b}}{\mathrm{2}}\:−\frac{{a}+\mathrm{2}{b}}{\mathrm{5}}\:\Rightarrow \\ $$$$\mathrm{3}\:=\mathrm{5}{a}+\mathrm{5}{b}\:−\mathrm{2}{a}\:−\mathrm{4}{b}\Rightarrow\mathrm{3}{a}\:+{b}=\mathrm{3} \\ $$$${F}\left(−\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\:=\frac{−{a}+{b}}{\mathrm{2}}\:−\frac{−{a}+\mathrm{2}{b}}{\mathrm{1}}\:\Rightarrow \\ $$$$−{a}+{b}\:+\mathrm{2}{a}\:−\mathrm{4}{b}\:=\mathrm{1}\:\Rightarrow{a}−\mathrm{3}{b}\:=\mathrm{1}\Rightarrow \\ $$$${a}=\mathrm{3}{b}\:+\mathrm{1}\:\Rightarrow\mathrm{3}\left(\mathrm{3}{b}+\mathrm{1}\right)\:+{b}\:=\mathrm{3}\:\Rightarrow \\ $$$$\mathrm{9}{b}\:+\mathrm{3}\:+{b}\:=\mathrm{3}\:\Rightarrow{b}=\mathrm{0}\:\Rightarrow{a}\:=\mathrm{1}\Rightarrow \\ $$$${F}\left({x}\right)\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{F}\left({x}\right){dx}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xdx}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{xdx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}} \\ $$$${but}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{xdx}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{x}+\mathrm{2}−\mathrm{2}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{5}\right)−{ln}\left(\mathrm{2}\right)\right\}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}\:+\mathrm{2}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=_{{x}+\mathrm{1}={u}} \:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:={arctan}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{4}}\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)\:=\:{arctan}\left(\mathrm{3}\right)−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$+\:{arctan}\left(\mathrm{2}\right)\:−\frac{\pi}{\mathrm{4}}\:\bigstar \\ $$