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Question Number 50219 by cesar.marval.larez@gmail.com last updated on 14/Dec/18

Answered by peter frank last updated on 15/Dec/18

$$allquestionabovelieontheconcept\left(partialfraction\right)$$$$\int \frac{5\mathrm{x}+3}{{\mathrm{x}}^3-2x^2-2\mathrm{x}}=\int \frac{5\mathrm{x}+3}{x(x+1)(x-3)}$$$$\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}-3}+\frac{\mathrm{C}}{\mathrm{x}+1}$$$$\mathrm{A}=-1\mathrm{B}=\frac{3}{2}\mathrm{C}=-\frac{1}{2}$$$$\int -\frac{1}{\mathrm{x}}\mathrm{dx}+\frac{3}{2}\int \frac{\mathrm{dx}}{\mathrm{x}-3}+-\frac{1}{2}\int \frac{\mathrm{dx}}{\mathrm{x}+1}$$$$-\ln \mathrm{x}+\frac{3}{2}\ln (x-3)-\frac{1}{2}\ln (x+1)+M$$$$$$

Answered by afachri last updated on 15/Dec/18

Answered by afachri last updated on 15/Dec/18

Commented by Abdo msup. last updated on 23/Dec/18

$$letdecomposeF\left(x\right)=\frac{3x^2-21x+32}{x{(x-4)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x-4}+\frac{c}{{(x-4)}^2}$$$$c={lim}_{x\to 4}{(x-4)}^{2}F\left(x\right)=\frac{3.16-21.4+32}{4}$$$$=12-21+8=-1\Rightarrow F\left(x\right)=\frac{a}{x}+\frac{b}{x-4}-\frac{1}{{(x-4)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=3=a+b$$$$F\left(1\right)=\frac{3-21+32}{9}=\frac{35-21}{9}=\frac{14}{9}=a-\frac{b}{3}-\frac{1}{9}\Rightarrow$$$$14=9a-3b-1\Rightarrow 9a-3b=15\Rightarrow 3a-b=5\Rightarrow$$$$3a-(3-a)=5\Rightarrow 4a=8\Rightarrow a=2andb=1\Rightarrow$$$$F\left(x\right)=\frac{2}{x}+\frac{1}{x-4}-\frac{1}{{(x-4)}^2}\Rightarrow$$$$\int F\left(x\right)dx=2ln\mid x\mid +ln\mid x-4\mid +\frac{1}{x-4}+c.$$

Commented by Abdo msup. last updated on 23/Dec/18

$$iseethatthedecompositioningivenansweris$$$$notcorrectbuttheresultiscorrect.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Dec/18

$$3)\int \frac{5x^2+12x+9}{x(x^2+3x+2)}dx$$$$\int \frac{5x^2+12x+9}{x(x+1)(x+2)}dx=\int \frac{a}{x}+\frac{b}{x+1}+\frac{c}{x+2}dx$$$$alnx+bln(x+1)+cln(x+2)+d$$$$nowcalculatinga,b,c$$$$5x^2+12x+9=a(x+1)(x+2)+bx(x+2)+cx(x+1)$$$$putx=0$$$$9=2aa=\frac{9}{2}$$$$putx+1=0$$$$5{(-1)}^2+12(-1)+9=b\times -1\times (-1+2)$$$$2=-bb=-2$$$$putx+2=0$$$$5{(-2)}^2+12(-2)+9=c\times -2\times (-2+1)$$$$5=2cc=\frac{5}{2}$$$$soabsweris$$$$\frac{9}{2}lnx-2ln(x+1)+\frac{5}{2}ln(x+2)+d$$

Commented by cesar.marval.larez@gmail.com last updated on 15/Dec/18

$$myproblemisthebeginningto$$$$descompose,thankumyfriend$$

Answered by afachri last updated on 15/Dec/18

Commented by Abdo msup. last updated on 23/Dec/18

$$letdecomposeF\left(x\right)=\frac{2x^2+x-4}{x(x-2)(x+1)}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x-2}+\frac{c}{x+1}$$$$a={lim}_{x\to 0}xF\left(x\right)=\frac{-4}{(-2)}=2$$$$b={lim}_{x\to 2}(x-2)F\left(x\right)=\frac{6}{6}=1$$$$c={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{-3}{3}=-1\Rightarrow$$$$F\left(x\right)=\frac{2}{x}+\frac{1}{x-2}-\frac{1}{(\mathit{x}+1)}\Rightarrow$$$$\int F\left(x\right)dx?=2ln\mid x\mid +ln\mid x-2\mid -ln\mid x+1\mid +c.$$

Answered by afachri last updated on 15/Dec/18

$$\left(5\right)\int \frac{8x+1}{x(x^2-6x+9)}dx=\int \frac{8x+1}{x{(x-3)}^2}dx$$$$=\int \frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x-3}+\frac{\mathrm{C}x+\mathrm{D}}{{(x-3)}^2}dx$$$$x=0;x=3$$$$\mathbf{so}\mathbf{A}\mathbf{and}\mathbf{B}\mathbf{are}:\mathrm{A}=\frac{8\left(0\right)+1}{{(0-3)}^2}=\frac{1}{9};\mathrm{B}=\frac{8\left(3\right)+1}{3}=\frac{25}{3}$$$$\mathbf{eq}\mathbf{will}\mathbf{be}:\int \frac{1}{9x}+\frac{25}{3(x-3)}+\frac{\mathrm{C}x+\mathrm{D}}{{(x-3)}^2}dx=\int \frac{8x+1}{x{(x-3)}^2}dx$$$$(8x+1)=\frac{1}{9}{(x-3)}^2+\frac{25}{3}x(x-3)+x(\mathrm{C}x+\mathrm{D})$$$$9(8x+1)=x^2-6x+9+25x^2-75x+\mathrm{C}{x}^2+\mathrm{D}x$$$$9(8x+1)=(1+25+\mathrm{C})x^2+(-6-75+\mathrm{D})x+9$$$$\mathbf{look}:$$$$(1+25+\mathrm{C})=0\Rightarrow \mathrm{C}=-26$$$$(-6-75+\mathrm{D})=72\Rightarrow \mathrm{D}=153\Rightarrow \mathrm{C}x+\mathrm{D}=-26x+153$$$$\mathbf{so}\mathbf{eq}:$$$$\int \frac{1}{9x}+\frac{25}{3(x-3)}+\frac{153-26x}{{(x-3)}^2}=\frac{\ln x}{9}+\frac{25\ln (x-3)}{3}+\int \frac{-26x+153}{{(x-3)}^2}dx$$$$\bullet \bullet \int \frac{-26x+153}{{(x-3)}^2}dxu=x-3$$$$\mathbf{let}u=x-3\Rightarrow du=dx$$$$-26u+75=-26x+78+75$$$$\int \frac{-26u+75}{u^2}du=\int -\frac{26}{u}+\frac{75}{u^2}du$$$$=-26\ln u-75u^{-1}$$$$=-26\ln (x-3)+\frac{75}{x-3}$$$$\mathbf{so}\mathbf{the}\mathbf{result}:$$$$=\frac{\ln x}{9}+\frac{25\ln (x-3)}{3}-26\ln (x-3)+\frac{75}{x-3}+C$$$$=\frac{\ln x}{9}-\frac{53\ln (x-3)}{3}+\frac{75}{x-3}+C$$$$$$

Commented by Abdo msup. last updated on 24/Dec/18

$$letI=\int \frac{8x+1}{x(x^2-6x+9)}dx\Rightarrow$$$$I=\int \frac{8x+1}{x{(x-3)}^2}dxletdevompose$$$$F\left(x\right)=\frac{8x+1}{x{(x-3)}^2}\Rightarrow F\left(x\right)=\frac{a}{x}+\frac{b}{x-3}+\frac{c}{{(x-3)}^2}$$$$a={lim}_{x\to 0}xF\left(x\right)=\frac{1}{9}$$$$c={lim}_{x\to 3}{(x-3)}^{2}F\left(x\right)=\frac{25}{3}\Rightarrow$$$$F\left(x\right)=\frac{1}{9x}+\frac{b}{x-3}+\frac{25}{3{(x-3)}^2}$$$$F\left(1\right)=\frac{9}{4}=\frac{1}{9}-\frac{b}{2}+\frac{25}{12}\Rightarrow 9=\frac{4}{9}-2b+\frac{25}{3}\Rightarrow$$$$2b=\frac{4}{9}+\frac{25}{3}-\frac{27}{3}=\frac{4}{9}-\frac{2}{3}=\frac{4-6}{9}=-\frac{2}{9}\Rightarrow b=-\frac{1}{9}$$$$F\left(x\right)=\frac{1}{9x}-\frac{1}{9(x-3)}+\frac{25}{3{(x-3)}^2}\Rightarrow$$$$\int F\left(x\right)dx=\frac{1}{9}ln\mid x\mid -\frac{1}{9}ln\mid x-3\mid -\frac{25}{3(x-3)}+c.$$

Answered by afachri last updated on 15/Dec/18

$$\left(6\right)\int \frac{5x+7}{x(x^2+4x+4)}dx=\int \frac{5x+7}{x{(x+2)}^2}dx$$$$=\int \frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{x+2}+\frac{\mathrm{C}x+\mathrm{D}}{{(x+2)}^2}dx$$$$\mathrm{to}\mathrm{define}\mathbf{A}\mathrm{and}\mathbf{B}:$$$$x=0;x+2=0$$$$x=-2$$$$\mathrm{A}=\frac{5\left(0\right)+7}{{(0+2)}^2}=\frac{7}{4};\mathrm{B}=\frac{5(-2)+7}{(-2)}=\frac{3}{2}$$$$\mathrm{So}\mathrm{the}\mathrm{eq}:$$$$\int \frac{7}{4x}+\frac{3}{2(x+2)}+\frac{\mathrm{C}x+\mathrm{D}}{{(x+2)}^2}dx=\int \frac{5x+7}{x{(x+2)}^2}dx$$$$\mathrm{to}\mathrm{define}\mathbf{C}\mathrm{and}\mathbf{D}:$$$$\frac{7{(x+2)}^2}{4}+\frac{3x(x+2)}{2}+x(\mathrm{C}x+\mathrm{D})=5x+7\mathbf{then}\mathbf{multiplied}\mathbf{by}4$$$$7{(x+2)}^2+6x(x+2)+4x(\mathrm{C}x+\mathrm{D})=20x+28$$$$(7+6+4\mathrm{C})x^2+(28+12+4\mathrm{D})x+\left(28\right)=20x+28$$$$\mathbf{observe}:$$$$(7+6+4\mathrm{C})x^2=0x^2\Rightarrow \mathrm{C}=-\frac{13}{4}$$$$(28+12+4\mathrm{D})x=20x\Rightarrow \mathrm{D}=-5$$$$\mathbf{so},\mathrm{C}x+\mathrm{D}=(-\frac{13}{4}x-5)\frac{4}{4}=\frac{-13x-20}{4}$$$$=-\frac{13x+20}{4}$$$$\mathbf{eq}\mathbf{will}\mathbf{be}:\int \frac{7}{4x}+\frac{3}{2(x+2)}-\frac{(13x+20)}{4{(x+2)}^2}dx$$$$=\frac{7\ln x}{4}+\frac{3\ln (x+2)}{2}-\frac{13\ln (x+2)}{4}+\frac{6}{4(x+2)}+C$$$$=\frac{7\ln x}{4}-\frac{7\ln (x+2)}{4}+\frac{3}{2(x+2)}+C$$$$=\frac{7}{4}\left[\ln \left(\frac{x}{x+2}\right)\right]+\frac{3}{2(x+2)}+C$$$$$$

Commented by cesar.marval.larez@gmail.com last updated on 15/Dec/18

$$withuiamlearningmore.ihavemaster$$$$othercountry$$

Commented by afachri last updated on 15/Dec/18

$$\mathrm{hehehe}{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{not}\mathrm{that}\mathrm{good}\mathrm{my}\mathrm{friend}.\mathrm{i}\mathrm{am}$$$$\mathrm{still}\mathrm{a}\mathrm{student}.\mathrm{we}\mathrm{can}\mathrm{share}\mathrm{and}\mathrm{cmplete}$$$$\mathrm{each}\mathrm{other}\mathrm{here}:)\mathrm{Glad}\mathrm{to}\mathrm{give}\mathrm{a}\mathrm{friend}$$$$\mathrm{a}\mathrm{hand}:)$$

Commented by afachri last updated on 15/Dec/18

$$\mathrm{waw}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{great}\mathrm{sir}.{\mathrm{i}}^{\prime }\mathrm{ll}\mathrm{text}\mathrm{you}.{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{Indonesian}.$$$$$$

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