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Question Number 65166 by Rio Michael last updated on 25/Jul/19

$$Giventhatf\left(x\right)=\frac{2}{x^2-1}$$$$a)Expressf\left(x\right)inpartialfraction.$$$$b.Evaluate{\int }_3^{5}f\left(x\right)dx$$

Commented by mathmax by abdo last updated on 26/Jul/19

$$a)wehavef\left(x\right)=\frac{2}{(x-1)(x+1)}=\frac{1}{x-1}-\frac{1}{x+1}$$$$b){\int }_3^{5}f\left(x\right)dx={\int }_3^5(\frac{1}{x-1}-\frac{1}{x+1})dx={[ln\mid x-1\mid -ln\mid x+1\mid ]}_3^5$$$$={[ln\mid \frac{x-1}{x+1}\mid ]}_3^5=ln\left(\frac{4}{6}\right)-ln\left(\frac{2}{4}\right)=ln\left(\frac{2}{3}\right)-ln\left(\frac{1}{2}\right)$$$$=ln\left(2\right)-ln\left(3\right)+ln2=2ln2-ln\left(3\right).$$

Answered by mr W last updated on 25/Jul/19

$$f\left(x\right)=\frac{2}{x^2-1}=\frac{1}{x-1}-\frac{1}{x+1}$$$${\int }_3^{5}f\left(x\right)dx$$$$={\int }_3^5(\frac{1}{x-1}-\frac{1}{x+1})dx$$$$={\left[\ln \frac{x-1}{x+1}\right]}_3^5$$$$=\ln \frac{5-1}{5+1}-\ln \frac{3-1}{3+1}$$$$=\ln \frac{2}{3}-\ln \frac{1}{2}$$$$=\ln \frac{4}{3}$$

Commented by Rio Michael last updated on 25/Jul/19

$$thanks$$

Commented by Rio Michael last updated on 25/Jul/19

$$anyexplanationonhow$$$$ln\frac{x-1}{x+1}sobtained$$

Commented by MJS last updated on 25/Jul/19

$$\int (x-1)dx=\ln (x-1)$$$$\int (x+1)dx=\ln (x+1)$$$$\ln a-\ln b=\ln \frac{a}{b}$$$${\mathrm{e}}^{\ln a-\ln b}={\mathrm{e}}^{\ln a}\times {\mathrm{e}}^{-\ln a}=\frac{{\mathrm{e}}^{\ln a}}{{\mathrm{e}}^{\ln b}}=\frac{a}{b}\Rightarrow$$$$\Rightarrow \ln {\mathrm{e}}^{\ln a-\ln b}=\ln a-\ln b=\ln \frac{a}{b}$$

Commented by mr W last updated on 25/Jul/19

$$\int (\frac{1}{x-1}-\frac{1}{x+1})dx$$$$=\int \frac{1}{x-1}dx-\int \frac{1}{x+1}dx$$$$=\int \frac{1}{x-1}d(x-1)-\int \frac{1}{x+1}d(x+1)$$$$=\ln (x-1)-\ln (x+1)$$$$=\ln \frac{x-1}{x+1}$$

Commented by Rio Michael last updated on 25/Jul/19

$$thanksyouguyssomuch$$

Answered by meme last updated on 25/Jul/19

$$a)f\left(x\right)=\frac{2}{(x-1)(x+1)}=\frac{1}{(x-1)}-\frac{1}{(x+1)}$$$$b){\int }_3^{5}f\left(x\right)dx={\int }_3^5\frac{1}{(x-1)}dx-{\int }_3^5\frac{1}{(x+1)}dx$$$$={\left[ln(x-1)\right]}_3^5-{\left[ln(x+1)\right]}_3^5$$$$=ln4-ln2-ln6+ln4$$$$=4ln2-ln2-ln6$$$$=3ln2-ln6$$$$=ln\frac{4}{3}$$

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