Question

Integrate the function w..r. to x : $\frac{1}{x(x-2)(x-4)}$

Answer 1

Solution:To integrate the function,we must use partial fraction method of Integration

$\frac{1}{x(x-2)(x-4)}= \frac{A}{x}+\frac{B}{x-2}+ \frac{C}{x-4}\\\\A=\frac{1}{(x-2)(x-4)} put x=0\\\\ A=\frac{-1}{8}\\\\B=\frac{1}{(x)(x-4)} put x=2\\\\\\ B=\frac{-1}{4}\\\\\\C=\frac{1}{(x)(x-2)} put x=4\\\\\\C=\frac{1}{8}\\\\\\\frac{1}{x(x-2)(x-4)} = \frac{-1}{8x}+ \frac{-1}{(4x-2)}+ \frac{1}{8(x-4)}\\\\\\\int\frac{-1}{8x} dx+\int \frac{-1}{4(x-2)} dx+ \int\frac{1}{8(x-4)}dx$

We know that $\int \frac{1}{x} dx= log x$

$\int\frac{-1}{8x} dx+\int \frac{-1}{4(x-2)} dx+ \int\frac{1}{8(x-4)}dx\\\\\\=\frac{-1}{8}log x-\frac{1}{4}log(x-2)+\frac{1}{8} log (x-4)+C($