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Question Number 20167 by tammi last updated on 23/Aug/17

please solve it  integrate with respect to x  ∫((5x−2)/(3x^2 +2x+1))

$${please}\:{solve}\:{it} \\ $$$${integrate}\:{with}\:{respect}\:{to}\:{x} \\ $$$$\int\frac{\mathrm{5}{x}−\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}} \\ $$

Answered by ajfour last updated on 23/Aug/17

=(1/3)∫(((5/2)(2x+2/3)−11/3)/((x+1/3)^2 +((√2)/3)^2 ))dx  =(1/3)×(5/2)∫((2(x+1/3))/((x+1/3)^2 +((√2)/3)^2 ))dx            −((11)/9)∫(dx/((x+1/3)^2 +((√2)/3)^2 ))  =(5/6)ln (x^2 +((2x)/3)+(1/3))−((11)/9)×(3/(√2))tan^(−1) (((3x+1)/(√2)))+C  =(5/6)ln (3x^2 +2x+1)−((11)/(3(√2)))tan^(−1) (((3x+1)/(√2)))+C ′ .

$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\left(\mathrm{5}/\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{2}/\mathrm{3}\right)−\mathrm{11}/\mathrm{3}}{\left({x}+\mathrm{1}/\mathrm{3}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}/\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{2}\left({x}+\mathrm{1}/\mathrm{3}\right)}{\left({x}+\mathrm{1}/\mathrm{3}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}/\mathrm{3}\right)^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{11}}{\mathrm{9}}\int\frac{{dx}}{\left({x}+\mathrm{1}/\mathrm{3}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}/\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\frac{\mathrm{2}{x}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{11}}{\mathrm{9}}×\frac{\mathrm{3}}{\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}{x}+\mathrm{1}}{\sqrt{\mathrm{2}}}\right)+{C} \\ $$$$=\frac{\mathrm{5}}{\mathrm{6}}\mathrm{ln}\:\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)−\frac{\mathrm{11}}{\mathrm{3}\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}{x}+\mathrm{1}}{\sqrt{\mathrm{2}}}\right)+{C}\:'\:. \\ $$

Commented by tammi last updated on 23/Aug/17

thanksss

$${thanksss} \\ $$

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