Math, asked by Anonymous, 1 year ago

please help me.....no spam

Attachments:

Answers

Answered by Anonymous
4
GB FB knock on shkjskallala
Attachments:
Answered by InesWalston
1

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


InesWalston: cheers, mate!
Similar questions