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Question Number 48643 by cesar.marval.larez@gmail.com last updated on 26/Nov/18

Commented by maxmathsup by imad last updated on 26/Nov/18

$$1)letI=\int \frac{3x+2}{(x-1)(x-2)(x^2+3)}dxletdecomposeF\left(x\right)=\frac{3x+2}{(x-1)(x-2)(x^2+3)}$$$$F\left(x\right)=\frac{a}{x-1}+\frac{b}{x-2}+\frac{cx+d}{x^2+3}$$$$a={lim}_{x\to 1}(x-1)F\left(x\right)=-\frac{5}{4}$$$$b={lim}_{x\to 2}(x-2)F\left(x\right)=\frac{8}{7}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b+c\Rightarrow c=\frac{5}{4}-\frac{8}{7}=\frac{35-32}{28}=\frac{3}{28}$$$$F\left(0\right)=\frac{1}{3}=-a-\frac{b}{2}+\frac{d}{3}=\frac{5}{4}-\frac{4}{7}+\frac{d}{3}\Rightarrow 1=\frac{15}{4}-\frac{12}{7}+d\Rightarrow$$$$1=\frac{105-48}{28}+d=\frac{57}{28}+d\Rightarrow d=1-\frac{57}{28}=-\frac{57-28}{28}=-\frac{99}{28}\Rightarrow$$$$F\left(x\right)=-\frac{5}{4(x-1)}+\frac{8}{7(x-2)}+\frac{\frac{3}{28}x-\frac{99}{28}}{x^2+3}\Rightarrow I=\int F\left(x\right)dx$$$$=-\frac{5}{4}ln\mid x-1\mid +\frac{8}{7}ln\mid x-2\mid +\frac{3}{56}\int \frac{2x}{x^2+3}dx-\frac{99}{28}\int \frac{dx}{x^2+3}$$$$=-\frac{5}{4}ln\mid x-1\mid +\frac{8}{7}ln\mid x-2\mid +\frac{3}{56}ln(x^2+3)-\frac{99}{28}\int \frac{dx}{x^2+3}but$$$$\int \frac{dx}{x^2+3}{=}_{x=\sqrt{3}t}\int \frac{1}{3(1+t^2)}\sqrt{3}dt=\frac{1}{\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\right)+c\Rightarrow$$$$I=-\frac{5}{4}ln\mid x-1\mid +\frac{8}{7}ln\mid x-2\mid +\frac{3}{56}ln(x^2+3)-\frac{99}{28\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\right)+C.$$$$$$

Commented by maxmathsup by imad last updated on 26/Nov/18

$$sorryerroroftypingd=-\frac{29}{28}\Rightarrow$$$$I=-\frac{5}{4}ln\mid x-1\mid +\frac{8}{7}ln\mid x-2\mid +\frac{3}{56}ln(x^2+3)-\frac{29}{28\sqrt{3}}arctan\left(\frac{x}{\sqrt{3}}\right)+C.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Nov/18

$$2)\int \frac{5x+3}{x^2+2}dx$$$$\frac{5}{2}\int \frac{2x}{x^2+2}dx+3\int \frac{dx}{x^2+2}$$$$\frac{5}{2}ln(x^2+2)+\frac{3}{\sqrt{2}}{tan}^{-1}\left(\frac{x}{\sqrt{2}}\right)+c$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

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