Question Number 7422 by Tawakalitu. last updated on 28/Aug/16
$$\int\frac{{x}^{\mathrm{2}} \:{dx}}{\left({x}\:−\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:\mathrm{4}\right)^{\mathrm{2}} } \\ $$
Answered by Yozzia last updated on 28/Aug/16
![(x^2 /((x−1)(x^2 +4)^2 ))≡(a/(x−1))+((bx+c)/(x^2 +4))+((ex+f)/((x^2 +4)^2 )) (x^2 /((x−1)(x^2 +4)^2 ))≡((a(x^2 +4)^2 +(bx+c)(x−1)(x^2 +4)+(ex+f)(x−1))/((x−1)(x^2 +4)^2 )) ∴ x^2 =a(x^2 +4)^2 +(bx+c)(x−1)(x^2 +4)+(ex+f)(x−1) x=1⇒1=a(5)^2 +0⇒a=(1/(25)) x=2i⇒−4=a(−4+4)^2 +0+(2ei+f)(2i−1) ∴f+2ei=((−4)/(2i−1))=((−4(−2i−1))/5) x=−2i⇒−4=0+0+(−2ie+f)(−2i−1) ∴ f−2ei=(4/(1+2i))=((4(1−2i))/5) ∴ 2f+0=((4(1−2i+2i+1))/5)=(8/5)⇒f=(4/5) ∴ (4/5)+2ei=(4/5)(1+2i)⇒e=(4/5) x=0⇒0=((16)/(25))+c(−1)(4)+(4/5)(0+1)(−1) 0=((16)/(25))−(4/5)−4c ⇒c=(4/(25))−(1/5)=((20−25)/(125))=((−1)/(25)) x=2⇒4=(1/(25))64+(2b−(1/(25)))(1)(8)+(4/5)(2+1)(1) 4=((64)/(25))+16b−(8/(25))+((60)/(25))=((116)/(25))+16b b=(1/(16))(4−((116)/(25)))=((−1)/(25)) ∴(x^2 /((x−1)(x^2 +4)^2 ))=(1/(25(x−1)))−((x+1)/(25(x^2 +4)))+((4(x+1))/(5(x^2 +4)^2 )) ∫(x^2 /((x−1)(x^2 +4)^2 ))dx=∫((1/(25(x−1)))−(x/(25(x^2 +4)))−(1/(25(x^2 +4)))+((4x)/(5(x^2 +4)^2 ))+(4/(5(x^2 +4)^2 )))dx ∫(x^2 /((x−1)(x^2 +4)^2 ))dx=∫(1/(25(x−1)))dx−(1/2)∫((2x)/(25(x^2 +4)))dx−∫(1/(25(x^2 +4)))dx+∫((4x)/(5(x^2 +4)^2 ))dx+∫(4/(5(x^2 +4)^2 ))dx −−−−−−−−−−−−−−−−−−−−−−−−−−−−− (1)∫(1/(25(x−1)))dx=(1/(25))ln∣x−1∣+c(1) (2)(1/(25))∫((2x)/(x^2 +4))dx=(1/(25))ln(x^2 +4)+c(2) (3)(1/(25))∫(1/(x^2 +4))dx=(1/(50))tan^(−1) (x/2)+c(3) (4)(2/5)∫((2x)/((x^2 +4)^2 ))dx [u=x^2 +4⇒du=2xdx] (2/5)∫((2x)/((x^2 +4)^2 ))dx=(2/5)∫(du/u^2 )=((−2)/(5u))+c(4)=((−2)/(5(x^2 +4)))+c(4) (5)∫(4/(5(x^2 +4)^2 ))dx [x=2tank⇒ dx=2sec^2 kdk] ∫(4/(5(x^2 +4)^2 ))dx=∫((8sec^2 kdk)/(5(4tan^2 k+4)^2 ))=(8/(80))∫((sec^2 kdk)/(sec^4 k))=(1/(10))∫cos^2 kdk=(1/(20))∫(1+cos2k)dk =(1/(20))(k+(1/2)sin2k) ∫(4/(5(x^2 +4)^2 ))dx=(1/(20))k+(1/(20))sinkcosk ∵ tank=(x/2)⇒sink=(x/( (√(x^2 +4)))), cosk=(2/( (√(x^2 +4)))) ∫(4/(5(x^2 +4)^2 ))dx=(1/(20))tan^(−1) (x/2)+(x/(10(x^2 +4)))+c(5) −−−−−−−−−−−−−−−−−−−−−−−−−−− ∴ ∫(x^2 /((x−1)(x^2 +4)^2 ))dx=(1/(25))ln∣x−1∣−(1/(50))ln(x^2 +4)+((50−20)/(1000))tan^(−1) (x/2)−(2/(5(x^2 +4)))+(x/(10(x^2 +4)))+C ∴ ∫(x^2 /((x−1)(x^2 +4)^2 ))dx=(1/(25))ln∣x−1∣−(1/(50))ln(x^2 +4)+(3/(100))tan^(−1) (x/2)+((x−4)/(10(x^2 +4)))+C](https://www.tinkutara.com/question/Q7430.png)
$$\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\equiv\frac{{a}}{{x}−\mathrm{1}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} +\mathrm{4}}+\frac{{ex}+{f}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\equiv\frac{{a}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} +\left({bx}+{c}\right)\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({ex}+{f}\right)\left({x}−\mathrm{1}\right)}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\therefore\:{x}^{\mathrm{2}} ={a}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} +\left({bx}+{c}\right)\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({ex}+{f}\right)\left({x}−\mathrm{1}\right) \\ $$$${x}=\mathrm{1}\Rightarrow\mathrm{1}={a}\left(\mathrm{5}\right)^{\mathrm{2}} +\mathrm{0}\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$${x}=\mathrm{2}{i}\Rightarrow−\mathrm{4}={a}\left(−\mathrm{4}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{0}+\left(\mathrm{2}{ei}+{f}\right)\left(\mathrm{2}{i}−\mathrm{1}\right) \\ $$$$\therefore{f}+\mathrm{2}{ei}=\frac{−\mathrm{4}}{\mathrm{2}{i}−\mathrm{1}}=\frac{−\mathrm{4}\left(−\mathrm{2}{i}−\mathrm{1}\right)}{\mathrm{5}} \\ $$$${x}=−\mathrm{2}{i}\Rightarrow−\mathrm{4}=\mathrm{0}+\mathrm{0}+\left(−\mathrm{2}{ie}+{f}\right)\left(−\mathrm{2}{i}−\mathrm{1}\right) \\ $$$$\therefore\:{f}−\mathrm{2}{ei}=\frac{\mathrm{4}}{\mathrm{1}+\mathrm{2}{i}}=\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}{i}\right)}{\mathrm{5}} \\ $$$$\therefore\:\mathrm{2}{f}+\mathrm{0}=\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}{i}+\mathrm{2}{i}+\mathrm{1}\right)}{\mathrm{5}}=\frac{\mathrm{8}}{\mathrm{5}}\Rightarrow{f}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\therefore\:\frac{\mathrm{4}}{\mathrm{5}}+\mathrm{2}{ei}=\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{1}+\mathrm{2}{i}\right)\Rightarrow{e}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${x}=\mathrm{0}\Rightarrow\mathrm{0}=\frac{\mathrm{16}}{\mathrm{25}}+{c}\left(−\mathrm{1}\right)\left(\mathrm{4}\right)+\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{0}+\mathrm{1}\right)\left(−\mathrm{1}\right) \\ $$$$\mathrm{0}=\frac{\mathrm{16}}{\mathrm{25}}−\frac{\mathrm{4}}{\mathrm{5}}−\mathrm{4}{c} \\ $$$$\Rightarrow{c}=\frac{\mathrm{4}}{\mathrm{25}}−\frac{\mathrm{1}}{\mathrm{5}}=\frac{\mathrm{20}−\mathrm{25}}{\mathrm{125}}=\frac{−\mathrm{1}}{\mathrm{25}} \\ $$$${x}=\mathrm{2}\Rightarrow\mathrm{4}=\frac{\mathrm{1}}{\mathrm{25}}\mathrm{64}+\left(\mathrm{2}{b}−\frac{\mathrm{1}}{\mathrm{25}}\right)\left(\mathrm{1}\right)\left(\mathrm{8}\right)+\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{2}+\mathrm{1}\right)\left(\mathrm{1}\right) \\ $$$$\mathrm{4}=\frac{\mathrm{64}}{\mathrm{25}}+\mathrm{16}{b}−\frac{\mathrm{8}}{\mathrm{25}}+\frac{\mathrm{60}}{\mathrm{25}}=\frac{\mathrm{116}}{\mathrm{25}}+\mathrm{16}{b} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{4}−\frac{\mathrm{116}}{\mathrm{25}}\right)=\frac{−\mathrm{1}}{\mathrm{25}} \\ $$$$\therefore\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{25}\left({x}−\mathrm{1}\right)}−\frac{{x}+\mathrm{1}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+\frac{\mathrm{4}\left({x}+\mathrm{1}\right)}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\int\left(\frac{\mathrm{1}}{\mathrm{25}\left({x}−\mathrm{1}\right)}−\frac{{x}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}−\frac{\mathrm{1}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+\frac{\mathrm{4}{x}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }+\frac{\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\right){dx} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\int\frac{\mathrm{1}}{\mathrm{25}\left({x}−\mathrm{1}\right)}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}−\int\frac{\mathrm{1}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}+\int\frac{\mathrm{4}{x}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}+\int\frac{\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left(\mathrm{1}\right)\int\frac{\mathrm{1}}{\mathrm{25}\left({x}−\mathrm{1}\right)}{dx}=\frac{\mathrm{1}}{\mathrm{25}}{ln}\mid{x}−\mathrm{1}\mid+{c}\left(\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\frac{\mathrm{1}}{\mathrm{25}}\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}=\frac{\mathrm{1}}{\mathrm{25}}{ln}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+{c}\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{3}\right)\frac{\mathrm{1}}{\mathrm{25}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}=\frac{\mathrm{1}}{\mathrm{50}}{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+{c}\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{4}\right)\frac{\mathrm{2}}{\mathrm{5}}\int\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\:\:\left[{u}={x}^{\mathrm{2}} +\mathrm{4}\Rightarrow{du}=\mathrm{2}{xdx}\right] \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}\int\frac{\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{du}}{{u}^{\mathrm{2}} }=\frac{−\mathrm{2}}{\mathrm{5}{u}}+{c}\left(\mathrm{4}\right)=\frac{−\mathrm{2}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+{c}\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{5}\right)\int\frac{\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\:\left[{x}=\mathrm{2}{tank}\Rightarrow\:{dx}=\mathrm{2}{sec}^{\mathrm{2}} {kdk}\right] \\ $$$$\int\frac{\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\int\frac{\mathrm{8}{sec}^{\mathrm{2}} {kdk}}{\mathrm{5}\left(\mathrm{4}{tan}^{\mathrm{2}} {k}+\mathrm{4}\right)^{\mathrm{2}} }=\frac{\mathrm{8}}{\mathrm{80}}\int\frac{{sec}^{\mathrm{2}} {kdk}}{{sec}^{\mathrm{4}} {k}}=\frac{\mathrm{1}}{\mathrm{10}}\int{cos}^{\mathrm{2}} {kdk}=\frac{\mathrm{1}}{\mathrm{20}}\int\left(\mathrm{1}+{cos}\mathrm{2}{k}\right){dk} \\ $$$$=\frac{\mathrm{1}}{\mathrm{20}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}{k}\right) \\ $$$$\int\frac{\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{20}}{k}+\frac{\mathrm{1}}{\mathrm{20}}{sinkcosk} \\ $$$$\because\:{tank}=\frac{{x}}{\mathrm{2}}\Rightarrow{sink}=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}},\:{cosk}=\frac{\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}} \\ $$$$\int\frac{\mathrm{4}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{20}}{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{10}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+{c}\left(\mathrm{5}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$$\therefore\:\int\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{25}}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{50}}{ln}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\frac{\mathrm{50}−\mathrm{20}}{\mathrm{1000}}{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+\frac{{x}}{\mathrm{10}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+{C} \\ $$$$\therefore\:\int\frac{{x}^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{1}}{\mathrm{25}}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{50}}{ln}\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\frac{\mathrm{3}}{\mathrm{100}}{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}+\frac{{x}−\mathrm{4}}{\mathrm{10}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}+{C} \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 28/Aug/16
$${Wow},\:{thanks}\:{so}\:{much}\:{sir}.\:{i}\:{really}\:{appreciate} \\ $$
Commented by Tawakalitu. last updated on 28/Aug/16
$${God}\:{bless}\:{you} \\ $$