Question
Evaluate: $\int\frac{\text{x}^{2}}{(\text{x}^{2} + 4)(\text{x}^{2} + 9 )}\text{dx}.$
✓
Answer
Let $\text{I} = \int\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )|(\text{x}^{2} + 9 )}\text{dx}$
Let $x^2 = t$
$\therefore\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )(\text{x}^{2} + 9 )} =\frac{\text{t}}{(\text{t} + 4)(\text{t} + 9 )}$
Now $\frac{\text{t}}{(\text{t} + 4 )(\text{t} + 9 )} =\frac{\text{A}}{\text{t} + 4} + \frac{\text{B}}{\text{t} + 9 } = \frac{\text{A}(\text{t} + 9) + \text{B}(\text{t} + 4)}{(\text{t} + 4 )(\text{t} + 9 )}$
$\Rightarrow$t =(A+ B) t +(9A+ 4B)
Equating we get
$A + B=1, 9A + 4B= 0$
Solving above two equations, we get
$\text{A} = - \frac{4}{5},\text{B} = \frac{9}{5}$
$\therefore\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )(\text{x}^{2} + 9 )} = -\frac{4}{5(\text{x}^{2} + 4 )} + \frac{9}{5(\text{x}^{2} + 9)}$
$\text{I} = -\frac{4}{5}\int\frac{\text{dx}}{\text{x}^{2} +2^{2}} +\frac{9}{5}\int\frac{\text{dx}}{\text{x}^{2} + 3^{2}}$
$ =-\frac{4}{5}\times\frac{1}{2}\tan^{-1} \frac{\text{x}}{2}+ \frac{9}{5}\times\frac{1}{3}\tan^{-1}\frac{\text{x}}{3} + \text{C}$
$ = - \frac{2}{5}\tan^{-1}\frac{\text{x}}{2} +\frac{3}{5}\tan^{-1}\frac{\text{x}}{3} + \text{C}.$
Let $x^2 = t$
$\therefore\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )(\text{x}^{2} + 9 )} =\frac{\text{t}}{(\text{t} + 4)(\text{t} + 9 )}$
Now $\frac{\text{t}}{(\text{t} + 4 )(\text{t} + 9 )} =\frac{\text{A}}{\text{t} + 4} + \frac{\text{B}}{\text{t} + 9 } = \frac{\text{A}(\text{t} + 9) + \text{B}(\text{t} + 4)}{(\text{t} + 4 )(\text{t} + 9 )}$
$\Rightarrow$t =(A+ B) t +(9A+ 4B)
Equating we get
$A + B=1, 9A + 4B= 0$
Solving above two equations, we get
$\text{A} = - \frac{4}{5},\text{B} = \frac{9}{5}$
$\therefore\frac{\text{x}^{2}}{(\text{x}^{2} + 4 )(\text{x}^{2} + 9 )} = -\frac{4}{5(\text{x}^{2} + 4 )} + \frac{9}{5(\text{x}^{2} + 9)}$
$\text{I} = -\frac{4}{5}\int\frac{\text{dx}}{\text{x}^{2} +2^{2}} +\frac{9}{5}\int\frac{\text{dx}}{\text{x}^{2} + 3^{2}}$
$ =-\frac{4}{5}\times\frac{1}{2}\tan^{-1} \frac{\text{x}}{2}+ \frac{9}{5}\times\frac{1}{3}\tan^{-1}\frac{\text{x}}{3} + \text{C}$
$ = - \frac{2}{5}\tan^{-1}\frac{\text{x}}{2} +\frac{3}{5}\tan^{-1}\frac{\text{x}}{3} + \text{C}.$
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