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Question Number 124406 by joki last updated on 03/Dec/20

Answered by Bird last updated on 03/Dec/20

$$letdecomposeF\left(x\right)=\frac{x+3}{{(x-1)}^2(x+4)}$$$$F\left(x\right)=\frac{a}{x-1}+\frac{b}{{(x-1)}^2}+\frac{c}{x+4}$$$$b={(x-1)}^{2}F\left(x\right){\mid }_{x=1}=\frac{4}{5}$$$$c=(x+4)F\left(x\right){\mid }_{x=-4}=\frac{-1}{25}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow a=\frac{1}{25}$$$$\Rightarrow F\left(x\right)=\frac{1}{25(x-1)}+\frac{4}{5{(x-1)}^2}-\frac{1}{25(x+4)}$$$$\Rightarrow \int F\left(x\right)dx=\frac{1}{25}ln\mid x-1\mid -\frac{4}{5(x-1)}$$$$-\frac{1}{25}ln\mid x+4\mid +C$$$$=\frac{1}{25}ln\mid \frac{x-1}{x+4}\mid -\frac{4}{5(x-1)}+C$$

Answered by MJS_new last updated on 03/Dec/20

$$\int \frac{x+3}{{(x-1)}^2(x+4)}dx=$$$$=\frac{4}{5}\int \frac{dx}{{(x-1)}^2}+\frac{1}{25}\int \frac{dx}{x-1}-\frac{1}{25}\int \frac{dx}{x+4}=$$$$=-\frac{4}{5(x-1)}+\frac{1}{25}\ln \mid \frac{x-1}{x+4}\mid +C$$

Commented by joki last updated on 03/Dec/20

$$idontunderstandsir,canyouexplainwith$$$$detailedsir.thankyou$$

Commented by MJS_new last updated on 03/Dec/20

$$\mathrm{just}\mathrm{decompose}\mathrm{the}\mathrm{fraction}$$

Commented by Ar Brandon last updated on 03/Dec/20

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Commented by Ar Brandon last updated on 03/Dec/20

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+3}{{(\mathrm{x}-1)}^2(\mathrm{x}+4)}=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{b}}{{(\mathrm{x}-1)}^2}+\frac{\mathrm{c}}{\mathrm{x}+4}$$$$=\frac{\mathrm{a}(\mathrm{x}-1)(\mathrm{x}+4)+\mathrm{b}(\mathrm{x}+4)+\mathrm{c}{(\mathrm{x}-1)}^2}{{(\mathrm{x}-1)}^2(\mathrm{x}+4)}$$$$\mathrm{x}=1\Rightarrow 4=5\mathrm{b},\mathrm{b}=\frac{4}{5}$$$$\mathrm{x}=-4\Rightarrow -1=25\mathrm{c},\mathrm{c}=-\frac{1}{25}$$$$\mathrm{for}\mathrm{coef}\mathrm{of}{\mathrm{x}}^2\mathrm{we}\mathrm{have}\mathrm{a}+\mathrm{c}=0,\mathrm{a}=\frac{1}{25}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{25(\mathrm{x}-1)}+\frac{4}{5{(\mathrm{x}-1)}^2}-\frac{1}{25(\mathrm{x}+4)}$$

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