Question
Evaluvate the following intregals:
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$
$\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}$
Put
x = 0 $\Rightarrow1=8\text{A}\Rightarrow\text{A}=\frac{1}{8}$ Put x = 2 $\Rightarrow1=-4\text{B}\Rightarrow\text{B}=-\frac{1}{4}$ Put x = 4 $\Rightarrow1=8\text{C}\Rightarrow\text{C}=\frac{1}{8}$ So, $\int\frac{1}{\text{x}(\text{x}-2)(\text{x}-4)}\ \text{dx}=\frac{1}{8}\int\frac{\text{dx}}{\text{x}}+\Big(-\frac{1}{4}\Big)\int\frac{\text{dx}}{\text{x}-2}+\frac{1}{8}\int\frac{\text{dx}}{\text{x}-4}$ $=\frac{1}{8}\log|\text{x}|-\frac{1}{4}\log|\text{x}-2|+\frac{1}{8}\log|\text{x}-4|+\text{C}$ $=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$ $\text{I}=\frac{1}{8}\log\Big|\frac{\text{x}(\text{x}-4)}{(\text{x}-2)^2}\Big|+\text{C}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(\text{y + xy})\text{dx}+(\text{x}-\text{xy}^2)\text{dy}=0$