Question Number 25483 by rita1608 last updated on 11/Dec/17
$${valute}\:\int\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\frac{}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx} \\ $$
Answered by prakash jain last updated on 11/Dec/17
![use partial fraction (1/((x−1)^2 (x^2 +4)))=(A/(x−1))+(B/((x−1)^2 ))+((Cx+D)/((x^2 +4))) A(x−1)(x^2 +4)+(B)(x^2 +4)+(Cx+D)(x−1)^2 =1 A(x^3 −x^2 +4x−4)+(B)(x^2 +4)+(Cx+D)(x^2 −2x+1)=1 x^3 : A+C=0⇒A=−C (I) x^2 : −A+B−2C+D=0 ⇒C+B−2C+D=0⇒B−C+D=0 (II+ x: 4A+C−2D=0⇒−3C−2D=0 (III) ⇒D=−((3C)/2) substituting D in II B−C−((3C)/2)=0⇒B=((5C)/2) x^0 : −4A+4B+D=1 −4(−C)+4(((5C)/2))−((3C)/2)=1 ⇒((8C+20C−3C)/2)=1⇒C=(2/(25)) A=−C=−(2/(25)) B=((5C)/2)=(1/5) D=−((3C)/2)=((−3)/(25)) (1/((x−1)^2 (x^2 +4)))=(A/(x−1))+(B/((x−1)^2 ))+((Cx+D)/((x^2 +4))) =((−2)/(25(x−1)))+(1/(5(x−1)^2 ))+((2x−3)/(25(x^2 +4))) ∫−(2/(25(x−1)))dx=−(2/(25))ln (x−1)+c_1 ∫(1/(5(x−1)^2 ))dx=−(1/5)∙(1/((x−1)))+c_2 ∫((2x−3)/(25(x^2 +4)))dx= =(1/(25))[∫((2x)/(x^2 +4))−∫(3/(x^2 +4))dx] =(1/(25))[∫((2x)/(x^2 +4))−(3/4)∫(1/(((x/2))^2 +1))dx] =(1/(25))[ln (x^2 +4)−(3/2)tan^(−1) (x/2)]+c_3 ANS −(2/(25))ln (x−1)−(1/(5(x−1)))+ (1/(25))[ln (x^2 +4)−(3/2)tan^(−1) (x/2)]+c](https://www.tinkutara.com/question/Q25489.png)
$$\mathrm{use}\:\mathrm{partial}\:\mathrm{fraction} \\ $$$$\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{Cx}+{D}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${A}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({B}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({Cx}+{D}\right)\left({x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${A}\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}\right)+\left({B}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\left({Cx}+{D}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{1} \\ $$$${x}^{\mathrm{3}} :\:{A}+{C}=\mathrm{0}\Rightarrow{A}=−{C}\:\left({I}\right) \\ $$$${x}^{\mathrm{2}} :\:−{A}+{B}−\mathrm{2}{C}+{D}=\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow{C}+{B}−\mathrm{2}{C}+{D}=\mathrm{0}\Rightarrow{B}−{C}+{D}=\mathrm{0}\:\left({II}+\right. \\ $$$${x}:\:\mathrm{4}{A}+{C}−\mathrm{2}{D}=\mathrm{0}\Rightarrow−\mathrm{3}{C}−\mathrm{2}{D}=\mathrm{0}\:\left({III}\right) \\ $$$$\:\:\:\:\:\Rightarrow{D}=−\frac{\mathrm{3}{C}}{\mathrm{2}} \\ $$$${substituting}\:{D}\:{in}\:{II} \\ $$$${B}−{C}−\frac{\mathrm{3}{C}}{\mathrm{2}}=\mathrm{0}\Rightarrow{B}=\frac{\mathrm{5}{C}}{\mathrm{2}} \\ $$$${x}^{\mathrm{0}} :\:−\mathrm{4}{A}+\mathrm{4}{B}+{D}=\mathrm{1} \\ $$$$−\mathrm{4}\left(−{C}\right)+\mathrm{4}\left(\frac{\mathrm{5}{C}}{\mathrm{2}}\right)−\frac{\mathrm{3}{C}}{\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{8}{C}+\mathrm{20}{C}−\mathrm{3}{C}}{\mathrm{2}}=\mathrm{1}\Rightarrow{C}=\frac{\mathrm{2}}{\mathrm{25}} \\ $$$${A}=−{C}=−\frac{\mathrm{2}}{\mathrm{25}} \\ $$$${B}=\frac{\mathrm{5}{C}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${D}=−\frac{\mathrm{3}{C}}{\mathrm{2}}=\frac{−\mathrm{3}}{\mathrm{25}} \\ $$$$\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{B}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{{Cx}+{D}}{\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{25}\left({x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{5}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$$\int−\frac{\mathrm{2}}{\mathrm{25}\left({x}−\mathrm{1}\right)}{dx}=−\frac{\mathrm{2}}{\mathrm{25}}\mathrm{ln}\:\left({x}−\mathrm{1}\right)+{c}_{\mathrm{1}} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{5}\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{5}}\centerdot\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)}+{c}_{\mathrm{2}} \\ $$$$\int\frac{\mathrm{2}{x}−\mathrm{3}}{\mathrm{25}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\left[\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{4}}−\int\frac{\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\left[\int\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{4}}\int\frac{\mathrm{1}}{\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}}{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{25}}\left[\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}\right)−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}\right]+{c}_{\mathrm{3}} \\ $$$${ANS} \\ $$$$−\frac{\mathrm{2}}{\mathrm{25}}\mathrm{ln}\:\left({x}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{5}\left({x}−\mathrm{1}\right)}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{25}}\left[\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}\right)−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{\mathrm{2}}\right]+{c} \\ $$