Question
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}$
$\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}$
We express
$\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+4}+\frac{\text{Cx}+\text{D}}{\text{x}^2+9}$
$\Rightarrow\text{x}^2=(\text{Ax}+\text{B})(\text{x}^2+9)+(\text{Cx}+\text{D})(\text{x}^2+4)$
Equating the coefficient of $x^3, x^2$, x and constants, we get
0 = A + C and 1 = B + D and 0 = 9A + 4C
and 0 = 9B + 4D or A = 0 and or A = 0 and
$\text{B}=-\frac{4}{5}\text{ and }\text{C}=0,\text{ D }=\frac{9}{5}$
$\therefore\text{I}=\int\bigg(\frac{-\frac{4}{5}}{\text{x}^2+4}+\frac{\frac{9}{5}}{\text{x}^2+9}\bigg)\text{dx}$
$=-\frac{4}{5}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{9}{5}\int\frac{1}{\text{x}^2+9}\ \text{dx}$
$=-\frac{4}{5}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{9}{5}\times\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
$=-\frac{2}{5}\tan^{-1}\frac{\text{x}}{2}+\frac{3}{5}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}=-\frac{2}{5}\tan^{-1}\frac{\text{x}}{2}+\frac{3}{5}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
We express
$\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}=\frac{\text{Ax}+\text{B}}{\text{x}^2+4}+\frac{\text{Cx}+\text{D}}{\text{x}^2+9}$
$\Rightarrow\text{x}^2=(\text{Ax}+\text{B})(\text{x}^2+9)+(\text{Cx}+\text{D})(\text{x}^2+4)$
Equating the coefficient of $x^3, x^2$, x and constants, we get
0 = A + C and 1 = B + D and 0 = 9A + 4C
and 0 = 9B + 4D or A = 0 and or A = 0 and
$\text{B}=-\frac{4}{5}\text{ and }\text{C}=0,\text{ D }=\frac{9}{5}$
$\therefore\text{I}=\int\bigg(\frac{-\frac{4}{5}}{\text{x}^2+4}+\frac{\frac{9}{5}}{\text{x}^2+9}\bigg)\text{dx}$
$=-\frac{4}{5}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{9}{5}\int\frac{1}{\text{x}^2+9}\ \text{dx}$
$=-\frac{4}{5}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{9}{5}\times\frac{1}{3}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
$=-\frac{2}{5}\tan^{-1}\frac{\text{x}}{2}+\frac{3}{5}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{x}^2+4)(\text{x}^2+9)}\ \text{dx}=-\frac{2}{5}\tan^{-1}\frac{\text{x}}{2}+\frac{3}{5}\tan^{-1}\frac{\text{x}}{3}+\text{C}$
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