Question

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

$\rm \bullet$ type - Hots
$\rm \bullet$ weightage :- 8 marks

answer accordingly!!!!​

Answer 1

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

AnSwEr:-

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

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

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

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

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

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firstly, let's solve $\sf I_1$

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

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

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

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

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

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

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now let's solve $\sf I_2$

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

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

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

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

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

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

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