Math, asked by HopelesslyRomantic, 3 months ago

evaluate \displaystyle\sf\int\dfrac{x^2}{x^4+x^2+1}\:dx

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\rm \bullet weightage :- 8 marks

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Answers

Answered by Anonymous
208

Before starting with it, note that in this type of integral, we can't use the method of partial fraction directly.

AnSwEr:-

Let \displaystyle\sf I = \int\dfrac{x^2}{x^4+x^2+1}\:dx

\displaystyle\sf = \dfrac{1}{2}\int\dfrac{2x^2}{x^4+x^2+1}\:dx

\displaystyle\sf\implies I = \dfrac{1}{2} \int \dfrac{x^2+1+x^2-1}{x^4+x^2+1}\:dx

\displaystyle\sf\implies I = \dfrac{1}{2} \int \dfrac{x^2+1}{x^4+x^2+1}\:dx\:+\:\dfrac{1}{2}\:\int\dfrac{x^2-1}{x^4+x^2+1}\:dx

\displaystyle\sf\implies I = \dfrac{1}{2}\:I_1\:+\:\:\dfrac{1}{2}\:I_2

\rm

firstly, let's solve \sf I_1

\rm

\displaystyle\sf\:I_1=\int\dfrac{x^2+1}{x^4+x^2+1}\:dx

on dividing numerator and denominator by \sf x^2 we get

\displaystyle\sf I_1=\int\dfrac{\left(\dfrac{x^2}{x^2} + \dfrac{1}{x^2}\right)}{\left(\dfrac{x^4}{x^2}+\dfrac{x^2}{x^2}+\dfrac{1}{x^2}\right)}\:dx

\displaystyle\sf = \int\dfrac{\left(1+\dfrac{1}{x^2}\right)}{\left(x^2+1+\dfrac{1}{x^2}\right)}\:dx

\displaystyle\sf = \int\dfrac{\left(1+\dfrac{1}{x^2}\right)}{\left(x-\dfrac{1}{x}\right)^2+2+1}\:dx

\displaystyle\sf =  \int\dfrac{\left(1+\dfrac{1}{x^2}\right)}{\left(x-\dfrac{1}{x}\right)^2+(\sqrt{3})^2}\:dx

  • now use substitution, put \sf x-\dfrac{1}{x}=t
  • \sf\implies \left(1+\dfrac{1}{x^2}\right)\:dx=dt

\displaystyle\sf\therefore\:I_1=\int\dfrac{dt}{t^2+(\sqrt{3})^2}

\displaystyle\sf = \dfrac{1}{\sqrt{3}} \: tan^{-1}\:\dfrac{t}{\sqrt{3}} \bf\:\:\left(\because\int\dfrac{dx}{a^2+x^2} = \dfrac{1}{a} \: tan^{-1}\left(\dfrac{x}{a}\right)\right)

\sf = \dfrac{1}{\sqrt{3}}\:tan^{-1}\:\dfrac{\left(x-\dfrac{1}{x}\right)}{\sqrt{3}}

\boxed{\sf = \dfrac{1}{\sqrt{3}}\:tan^{-1}\:\left(\dfrac{x^2-1}{x\sqrt{3}}\right)}

\rm

now let's solve \sf I_2

\rm

\displaystyle\sf I_2 = \int\dfrac{x^2-1}{x^4+x^2+1}\:dx

on dividing numerator and denominator by \sf x^2 we get

\displaystyle\sf I_1=\int\dfrac{\left(\dfrac{x^2}{x^2} - \dfrac{1}{x^2}\right)}{\left(\dfrac{x^4}{x^2}+\dfrac{x^2}{x^2}+\dfrac{1}{x^2}\right)}\:dx

\displaystyle\sf = \int\dfrac{\left(1-\dfrac{1}{x^2}\right)}{\left(x^2+1+\dfrac{1}{x^2}\right)}\:dx

\displaystyle\sf = \int\dfrac{\left(1-\dfrac{1}{x^2}\right)}{\left(x+\dfrac{1}{x}\right)^2-2+1}\:dx

\displaystyle\sf =  \int\dfrac{\left(1-\dfrac{1}{x^2}\right)}{\left(x+\dfrac{1}{x}\right)^2-1}\:dx

now use substitution, put \sf x+\dfrac{1}{x}=t

\sf\implies \left(1-\dfrac{1}{x^2}\right)\:dx=dt

\displaystyle\sf I_2 = \int\dfrac{dt}{t^2-1}

\displaystyle\sf = \int\dfrac{dt}{t^2-(1)^2}

\displaystyle\sf = \dfrac{1}{2\times\:1} \: log \: \left|\dfrac{t-1}{t+1}\right|\:\:\bf \left(\because \: \int\dfrac{dx}{x^2-a^2}=\dfrac{1}{2a}\:log\:\left|\dfrac{x-a}{x+a}\right|\right)

\displaystyle\sf = \dfrac{1}{2}\:log\:\left|\dfrac{t-1}{t+1}\right|

  • put the value of t

\sf = \dfrac{1}{2}\:log\:\left|\dfrac{\left(x+\dfrac{1}{x}\right)-1}{\left(x+\dfrac{1}{x}\right)+1}\right|

\boxed{\sf = \dfrac{1}{2}\;log\:\left|\dfrac{x^2+1-x}{x^2+1+x}\right|}

\rm

now, on putting values of \sf I_1 & \sf I_2 , we get

\rm

\sf I = \dfrac{1}{2}\times\dfrac{1}{\sqrt{3}}\:tan^{-1}\:\left(\dfrac{x^2-1}{x\sqrt{3}}\right) + \dfrac{1}{2}\cdot\dfrac{1}{2}\:log\:\left|\dfrac{x^2-x+1}{x^2+x+1}\right|\:+\:C

\boxed{\sf \therefore I = \dfrac{1}{2\sqrt{3}}\:tan^{-1}\:\left(\dfrac{x^2-1}{x\sqrt{3}}\right)+\dfrac{1}{4}\:log\:\left|\dfrac{x^2-x+1}{x^2+x+1}\right|\:+\:C}


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