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

Question

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

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Answer

$\int\Big(\frac{\text{x}-3}{\text{x}^2+2\text{x}-4}\Big)\text{dx}$
$\text{x}-3=\text{A}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+2\text{x}-4\big)+\text{B}$
$\text{x}-3=\text{A}(2\text{x}+2)+\text{B}$
$\text{x}-3=(2\text{A})\text{x}+2\text{A}+\text{B}$
Comparing the coefficients of like power of x,
$2\text{A}=1$
$\text{A}=\frac{1}{2}$
$2\text{A}+\text{B}=-3$
$2\times\frac{1}{2}+\text{B}=-3$
$\text{B}=-4$
Now, $\int\Big(\frac{\text{x}-3}{\text{x}^2+2\text{x}-4}\Big)\text{dx}$
$=\int\bigg(\frac{\frac{1}{2}(2\text{x}+2)-4}{\text{x}^2+2\text{x}-4}\bigg)\text{dx}$
$=\frac{1}{2}\int\frac{(2\text{x}+2)\text{dx}}{(\text{x}^2+2\text{x}-4)}-4\int\frac{\text{dx}}{\text{x}^2+2\text{x}+1-1-4}$
$=\frac{1}{2}\int\frac{(2\text{x}+2)\text{dx}}{(\text{x}^2+2\text{x}-4)}-4\int\frac{\text{dx}}{(\text{x}+1)^2-(\sqrt5)^2}$
$=\frac{1}{2}\log\big|\text{x}^2+2\text{x}+4\big|-\frac{4}{2\sqrt5}\log\Big|\frac{\text{x}+1-\sqrt5}{\text{x}+1+\sqrt5}\Big|+\text{C}$
$=\frac{1}{2}\log\big|\text{x}^2+2\text{x}+4\big|-\frac{2}{\sqrt5}\log\Big|\frac{\text{x}+1-\sqrt5}{\text{x}+1+\sqrt5}\Big|+\text{C}$

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