Evaluate the following intregals: x^2+1 (x^2+4)(x^2+25) dx - Vidyadip

Question

Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$

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Answer

Consider the integrals
$\text{I}=\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
Let y = x²
Thus,
$\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}=\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}}{\text{y}+4}+\frac{\text{B}}{\text{y}+25}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}(\text{y}+25)+\text{B}(\text{y}+4)}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\text{y}+1=\text{Ay}+25\text{A}+\text{By}+4\text{B}$
Compairing the coefficient, we have,
A + B = 1 and 25A + 4B = 1
Solving the above equations, we have,
$\text{A}=\frac{-1}{7}\text{ and }\text{B}=\frac{8}{7}$
Thus, $\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
$=\int\frac{\frac{1}{7}}{\text{x}^2+4}\ \text{dx}+\int\frac{\frac{8}{7}}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{8}{7}\int\frac{1}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{7}\times\frac{1}{5}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
$=\frac{-1}{14}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{35}\tan^{-1}\frac{\text{x}}{5}+\text{C}$

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