Question

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Answer 1

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Answer 2

Solution-

$I=\int \dfrac{x^2+x+1}{x^2(x+2)} \, dx$

$=\int [\dfrac{x^2}{x^2(x+2)}+\dfrac{x}{x^2(x+2)}+\dfrac{1}{x^2(x+2)}] \, dx$

$=\int \dfrac{1}{(x+2)} \, dx+\int \dfrac{1}{x(x+2)} \, dx+\int \dfrac{1}{x^2(x+2)} \, dx$

$=I_1+I_2+I_3$

$I_1=\int \dfrac{1}{(x+2)} \, dx=\log (x+2)+c_1$

$I_2=\int \dfrac{1}{x(x+2)} \, dx$

$=\dfrac{1}{2}\int \dfrac{(x+2)-(x)}{x(x+2)} \, dx$

$=\dfrac{1}{2}[\int \dfrac{1}{x} \, dx-\int \dfrac{1}{x+2} \, dx]$

$=\dfrac{1}{2}[\log (x)-\log (x+2)]+c_2$

$I_3=\int \dfrac{1}{x^2(x+2)} \, dx$

Decomposing the fraction using partial fraction,

$\Rightarrow \dfrac{1}{x^2\left(x+2\right)}=\dfrac{a_0}{x}+\dfrac{a_1}{x^2}+\dfrac{a_2}{x+2}$

$\Rightarrow \dfrac{1\cdot \:x^2\left(x+2\right)}{x^2\left(x+2\right)}=\dfrac{a_0x^2\left(x+2\right)}{x}+\dfrac{a_1x^2\left(x+2\right)}{x^2}+\dfrac{a_2x^2\left(x+2\right)}{x+2}$

$\Rightarrow 1=a_0x\left(x+2\right)+a_1\left(x+2\right)+a_2x^2$

$\mathrm{For\:the\:denominator\:root}\:0:\quad a_1=\dfrac{1}{2}$

As,

$\Rightarrow 1=a_0\cdot \:0\cdot \left(0+2\right)+a_1\left(0+2\right)+a_2\cdot \:0^2\\\\\Rightarrow 1=2a_1\\\\\Rightarrow a_1=\frac{1}{2}$

$\mathrm{For\:the\:denominator\:root}\:-2:\quad a_2=\dfrac{1}{4}$

As,

$\Rightarrow 1=a_0\left(-2\right)\left(\left(-2\right)+2\right)+a_1\left(\left(-2\right)+2\right)+a_2\left(-2\right)^2\\\\\Rightarrow 1=4a_2\\\\\Rightarrow a_2=\dfrac{1}{4}$

Now putting these values in the equation,

$\Rightarrow 1=a_0x\left(x+2\right)+\dfrac{1}{2}\left(x+2\right)+\frac{1}{4}x^2\\\\\Rightarrow 1=a_0x^2+2a_0x+\dfrac{x}{2}+1+\dfrac{x^2}{4}$

$\Rightarrow 1=x^2\left(a_0+\dfrac{1}{4}\right)+x\left(2a_0+\dfrac{1}{2}\right)+1$

$\mathrm{Solve}\:2a_0+\frac{1}{2}=0\:\mathrm{for}\:a_0$

$a_0=-\dfrac{1}{4}$

$\mathrm{Plug\:the\:solutions\:to\:the\:partial\:fraction\:parameters\:to\:obtain\:the\:final\:result}$

$=\dfrac{\left(-\frac{1}{4}\right)}{x}+\dfrac{\frac{1}{2}}{x^2}+\dfrac{\frac{1}{4}}{x+2}$

$=\dfrac{1}{2x^2}+\dfrac{1}{4\left(x+2\right)}-\dfrac{1}{4x}$

Now,

$I_3=\int \dfrac{1}{2x^2}dx+\int \dfrac{1}{4\left(x+2\right)}dx-\int \dfrac{1}{4x}dx$

$=-\dfrac{1}{2x}+\dfrac{1}{4}\log (x+2)-\dfrac{1}{4}\log (x)+c_3$

$I=\log (x+2)+\dfrac{1}{2}\log (x)-\dfrac{1}{2}\log (x+2)-\dfrac{1}{2x}+\dfrac{1}{4}\log (x+2)-\dfrac{1}{4}\log (x)+c$

$=\dfrac{3}{4}\log (x+2)+\dfrac{1}{4}\log (x)-\dfrac{1}{2x}+c$