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

Question

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

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Answer

Consider the integral
$\text{I}=\int\frac{2\text{x}^2+1}{\text{x}^2(\text{x}^2+4)}\ \text{dx}$
Now let us seoarate the fraction $\frac{2\text{x}^2+1}{\text{x}^2(\text{x}^2+4)}$ through partial fractions.
Substitute $x^2 = t$, then
$\frac{2\text{x}^2+1}{\text{x}^2(\text{x}^2+4)}=\frac{2\text{t}+1}{\text{t}(\text{t}+4)}$
$\Rightarrow\frac{2\text{t}+1}{\text{t}(\text{t}+4)}=\frac{\text{A}}{\text{t}}+\frac{\text{B}}{\text{t}+4}$
$\Rightarrow\frac{2\text{t}+1}{\text{t}(\text{t}+4)}=\frac{\text{A}(\text{t}+4)+\text{Bt}}{\text{t}(\text{t}+4)}$
$=2\text{t}+1=\text{A}(\text{t}+4)+\text{Bt}$
$=2\text{t}+1=\text{At}+4\text{A}+\text{Bt}$
Compairing the coefficient, we have,
A + B = 2 and 4A = 1
$\Rightarrow\text{A}=\frac{1}{2}$ and $\text{B}=\frac{7}{4}$
$\Rightarrow\frac{2\text{x}^2+1}{\text{x}^2(\text{x}^2+4)}=\frac{1}{4\text{x}^2}+\frac{7}{4(\text{x}^2+4)}$
Thus, we have,
$\text{I}=\int\frac{2\text{x}^2+1}{(\text{x}^2+4)}\ \text{dx}$
$=\frac{1}{4}\int\frac{\text{dx}}{\text{x}^2}+\frac{7}{4}\int\frac{\text{dx}}{(\text{x}^2+4)}$
$=-\frac{1}{4\text{x}}+\frac{7}{4}\times\frac{1}{2}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
$=-\frac{1}{4\text{x}}+\frac{7}{8}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$

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