Evaluate the following integrals: x^2 (x^2+1)(3x^2+4) dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}\ \text{dx}$

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Answer

We have,
$\text{I}=\int\frac{\text{x}^2\text{dx}}{(\text{x}^2+1)(3\text{x}^2+4)} $
Putting $x^2 = t$
Then, $\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}$
Let $\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}=\frac{\text{A}}{\text{t}+1}+\frac{\text{B}}{3\text{t}+4}$
$\Rightarrow\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}=\frac{\text{A}(3\text{t}+4)+\text{B}(\text{t}+1)}{(\text{t}+1)(3\text{t}+4)}$
$\Rightarrow\text{t}=\text{A}(3\text{t}+4)+\text{B}(\text{t}+1)$
putting $t + 1 = 0$
$⇒ t = -1$
$\therefore$ -$1 = A (-3 + 4) + 0$
$⇒ A = -1$
putting $3t + 4 = 0$
$\Rightarrow\text{t}=-\frac{4}{3}$
$\therefore-\frac{4}{3}=0+\text{B}\big(-\frac{4}3{}+1\big)$
$\Rightarrow-\frac{4}{3}=\text{B}\times\big(-\frac{1}{3}\big)$
$\Rightarrow\text{B}=4$
$\therefore\frac{\text{t}}{(\text{t}+1)(3\text{t}+4)}=-\frac{1}{\text{t}+1}+\frac{4}{3\text{t}+4}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=\frac{1}{\text{x}^2+1}+\frac{4}{3\text{x}^2+4}$
$$$\Rightarrow\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=\frac{1}{\text{x}^2+1}+\frac{4}{3\big(\text{x}^2+\frac{4}{3}\big)}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2+1)(3\text{x}^2+4)}=-\int\frac{\text{dx}}{\text{x}^2+1}+\frac{4}{3}\int\frac{\text{dx}}{\text{x}^2+\big(\frac{2}{\sqrt{3}}\big)^2}$
$=-\tan^{-1}(\text{x})+\frac{4}{3}\times\frac{\sqrt{3}}{2}\tan^{-1}\big(\frac{\sqrt{3}\text{x}}{2}\big)+\text{C}$
$=-\tan^{-1}(\text{x})+\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{\sqrt{3}\text{x}}{2}\Big)+\text{C}$

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