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Question Number 20541 by tammi last updated on 28/Aug/17

∫((xdx)/((x−1)(x^2 +1)))

$$\int\frac{{xdx}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$

Answered by $@ty@m last updated on 29/Aug/17

Let (x/((x−1)(x^2 +1)))=(A/(x−1))+((Bx+C)/(x^2 +1))  ⇒A(x^2 +1)+Bx(x−1)+C(x−1)=x  ⇒A+B=0 , A−B+C=1 & A−C=0  ⇒A=C=(1/3) & B=((−1)/3)  Now try yourself....

$${Let}\:\frac{{x}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{A}}{{x}−\mathrm{1}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{A}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{Bx}\left({x}−\mathrm{1}\right)+{C}\left({x}−\mathrm{1}\right)={x} \\ $$$$\Rightarrow{A}+{B}=\mathrm{0}\:,\:{A}−{B}+{C}=\mathrm{1}\:\&\:{A}−{C}=\mathrm{0} \\ $$$$\Rightarrow{A}={C}=\frac{\mathrm{1}}{\mathrm{3}}\:\&\:{B}=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${Now}\:{try}\:{yourself}.... \\ $$$$\:\:\:\:\:\: \\ $$

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