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Question Number 87185 by M±th+et£s last updated on 03/Apr/20

∫((2x−1)/(x(x^2 +3)))dx

$$\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dx} \\ $$

Answered by peter frank last updated on 03/Apr/20

∫((2x)/(x(x^2 +3)))dx−∫(1/(x(x^2 +3)))dx ∫(2/((x^2 +3)))dx−∫(1/(x(x^2 +3)))dx

$$\int\frac{\mathrm{2}{x}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dx}−\int\frac{\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dx} \\ $$$$\int\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dx}−\int\frac{\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dx} \\ $$$$ \\ $$

Answered by Kunal12588 last updated on 03/Apr/20

((2x−1)/(x(x^2 +3)))=(a/x)+((bx+c)/(x^2 +3)) ⇒2x−1=(a+b)x^2 +cx+3a a=−(1/3), b=(1/3), c=2 I=−(1/3)∫(dx/x)+(1/3)∫((x dx)/(x^2 +3))+2∫(dx/(x^2 +3)) =−(1/3)ln ∣x∣+(1/6)ln ∣x^2 +3∣+(2/(√3))tan^(−1) (x/(√3))+C

$$\frac{\mathrm{2}{x}−\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}=\frac{{a}}{{x}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{x}−\mathrm{1}=\left({a}+{b}\right){x}^{\mathrm{2}} +{cx}+\mathrm{3}{a} \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{3}},\:{b}=\frac{\mathrm{1}}{\mathrm{3}},\:{c}=\mathrm{2} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{x}\:{dx}}{{x}^{\mathrm{2}} +\mathrm{3}}+\mathrm{2}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid{x}\mid+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\:\mid{x}^{\mathrm{2}} +\mathrm{3}\mid+\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \frac{{x}}{\sqrt{\mathrm{3}}}+{C} \\ $$

Commented by M±th+et£s last updated on 03/Apr/20

thank uou sir

$${thank}\:{uou}\:{sir} \\ $$

Answered by redmiiuser last updated on 03/Apr/20

∫(2/(x^2 +3))dx =(2/(√3))tan^(−1) ((x/((√)3)))+C ∫(dx/(x(x^2 +3))) x=(√3)tan z dx=(√3).sec^2 z.dz ∫(((√3).sec^2 z.dz)/((√3)tan z.3sec^2 z)) =(1/3)∫cotz.dz =(1/3)log ∣sin z∣+A =(1/3)log ∣sin (tan^(−1) ((x/(√3))))∣+A ∫((2x−1)/(x(x^2 +3))).dx =(2/(√3))tan^(−1) ((x/((√)3)))−(1/3)log ∣sin (tan^(−1) ((x/(√3))))∣+c

$$\int\frac{\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{}\mathrm{3}}\right)+{C} \\ $$$$\int\frac{{dx}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)} \\ $$$${x}=\sqrt{\mathrm{3}}\mathrm{tan}\:{z} \\ $$$${dx}=\sqrt{\mathrm{3}}.\mathrm{sec}\:^{\mathrm{2}} {z}.{dz} \\ $$$$\int\frac{\sqrt{\mathrm{3}}.{sec}^{\mathrm{2}} {z}.{dz}}{\sqrt{\mathrm{3}}\mathrm{tan}\:{z}.\mathrm{3sec}\:^{\mathrm{2}} {z}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int{cotz}.{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\:\mid\mathrm{sin}\:{z}\mid+{A} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\:\mid\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)\right)\mid+{A} \\ $$$$\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}.{dx} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{}\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}\:\mid\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)\right)\mid+{c} \\ $$$$ \\ $$

Commented by redmiiuser last updated on 03/Apr/20

PLS CHECK

$${PLS}\:{CHECK} \\ $$

Commented by Joel578 last updated on 03/Apr/20

you can check by differentiating it again

$${you}\:{can}\:{check}\:{by}\:{differentiating}\:{it}\:{again} \\ $$

Commented by M±th+et£s last updated on 03/Apr/20

nice solution sir

$${nice}\:{solution}\:{sir} \\ $$

Commented by M±th+et£s last updated on 03/Apr/20

its didinte integeral if he derivative it the answer will be zero

$${its}\:{didinte}\:{integeral}\:{if}\:{he}\:{derivative}\:{it}\:{the} \\ $$$${answer}\:{will}\:{be}\:{zero} \\ $$

Commented by M±th+et£s last updated on 03/Apr/20

∗∗∗∗difine integeral∗∗∗∗

$$\ast\ast\ast\ast{difine}\:{integeral}\ast\ast\ast\ast \\ $$

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