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Question Number 119335 by physicstutes last updated on 23/Oct/20

$$\mathrm{Express}f\left(x\right)=\frac{1}{{(x-1)}^2(x^2+1)}\mathrm{into}\mathrm{partial}\mathrm{fractions}.$$$$\mathrm{hence}\mathrm{evaluate}I=\underset{0}{\stackrel{4}{\int }}f\left(x\right)dx$$

Answered by floor(10²Eta[1]) last updated on 23/Oct/20

$$\frac{1}{{(\mathrm{x}-1)}^2({\mathrm{x}}^2+1)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{{(\mathrm{x}-1)}^2}+\frac{\mathrm{Cx}+\mathrm{D}}{{\mathrm{x}}^2+1}$$$$1=\mathrm{A}(\mathrm{x}-1)({\mathrm{x}}^2+1)+\mathrm{B}({\mathrm{x}}^2+1)+(\mathrm{Cx}+\mathrm{D}){(\mathrm{x}-1)}^2$$$$=(\mathrm{A}+\mathrm{C}){\mathrm{x}}^3-(\mathrm{A}-\mathrm{B}+2\mathrm{C}+\mathrm{D}){\mathrm{x}}^2+(\mathrm{A}+\mathrm{C}-2\mathrm{D})\mathrm{x}-\mathrm{A}+\mathrm{B}+\mathrm{D}$$$$\Rightarrow \mathrm{A}+\mathrm{C}=0$$$$\mathrm{A}-\mathrm{B}+2\mathrm{C}+\mathrm{D}=0$$$$\mathrm{A}+\mathrm{C}-2\mathrm{D}=0$$$$-\mathrm{A}+\mathrm{B}+\mathrm{D}=1$$$$\Rightarrow \mathrm{D}=0\Rightarrow \mathrm{C}=1/2\Rightarrow \mathrm{A}=-1/2\Rightarrow \mathrm{B}=1/2$$$$\frac{1}{{(\mathrm{x}-1)}^2({\mathrm{x}}^2+1)}=\frac{-1}{2(\mathrm{x}-1)}+\frac{1}{2{(\mathrm{x}-1)}^2}+\frac{\mathrm{x}}{2({\mathrm{x}}^2+1)}$$$$$$$$$$

Answered by mathmax by abdo last updated on 24/Oct/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{(\mathrm{x}-1)}^2({\mathrm{x}}^2+1)}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{b}}{{(\mathrm{x}-1)}^2}+\frac{\mathrm{cx}+\mathrm{d}}{{\mathrm{x}}^2+1}$$$$\mathrm{b}=\frac{1}{2},{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xf}\left(\mathrm{x}\right)=0=\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=-\mathrm{a}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{1}{2{(\mathrm{x}-1)}^2}+\frac{-\mathrm{ax}+\mathrm{d}}{{\mathrm{x}}^2+1}$$$$\mathrm{f}\left(0\right)=1=-\mathrm{a}+\frac{1}{2}+\mathrm{d}\Rightarrow -\mathrm{a}+\mathrm{d}=\frac{1}{2}\Rightarrow \mathrm{a}-\mathrm{d}=-\frac{1}{2}$$$$\mathrm{f}\left(2\right)=\frac{1}{5}=\mathrm{a}+\frac{1}{2}+\frac{-2\mathrm{a}+\mathrm{d}}{5}\Rightarrow 1=5\mathrm{a}+\frac{5}{2}-2\mathrm{a}+\mathrm{d}\Rightarrow$$$$3\mathrm{a}+\mathrm{d}=1-\frac{5}{2}=-\frac{3}{2}\mathrm{we}\mathrm{have}\mathrm{a}=\mathrm{d}-\frac{1}{2}\Rightarrow 3(\mathrm{d}-\frac{1}{2})+\mathrm{d}=-\frac{3}{2}$$$$\Rightarrow 4\mathrm{d}=0\Rightarrow \mathrm{d}=0\Rightarrow \mathrm{a}=-\frac{1}{2}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=-\frac{1}{2(\mathrm{x}-1)}+\frac{1}{2{(\mathrm{x}-1)}^2}+\frac{\mathrm{x}}{2({\mathrm{x}}^2+1)}\mathrm{so}$$$${\int }_0^4\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=-\frac{1}{2}{\int }_0^4\frac{\mathrm{dx}}{\mathrm{x}-1}+\frac{1}{2}{\int }_0^4\frac{\mathrm{dx}}{{(\mathrm{x}-1)}^2}+\frac{1}{4}{\int }_0^4\frac{2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}$$$$=-\frac{1}{2}{[\ln \mid \mathrm{x}-1\mid ]}_0^4-\frac{1}{2}{\left[\frac{1}{\mathrm{x}-1}\right]}_0^4+\frac{1}{4}{\left[\ln ({\mathrm{x}}^2+1)\right]}_0^4$$$$=-\frac{1}{2}\ln \left(3\right)-\frac{1}{2}(\frac{1}{3}+1)+\frac{1}{4}\ln \left(17\right)$$$$=-\frac{\ln 3}{2}-\frac{2}{3}+\frac{\ln 17}{4}$$

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