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The integral value of \int\frac{x-1}{\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )}dx
 

Option: 1

\frac{1}{2}log\left | x-1 \right |-2log\left | x-3 \right |+\frac{3}{2}log\left | x-4 \right |+C


Option: 2

\frac{1}{2}log\left | x-1 \right |+2log\left | x-3 \right |+\frac{3}{2}log\left | x-4 \right |+C


Option: 3

\frac{1}{2}log\left | x-1 \right |-2log\left | x-3 \right |-\frac{3}{2}log\left | x-4 \right |+C


Option: 4

-\frac{1}{2}log\left | x-1 \right |-2log\left | x-3 \right |-\frac{3}{2}log\left | x-4 \right |+C


Answers (1)

best_answer

Given integral,

\int\frac{x-1}{\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )}dx

\int\frac{x-1}{\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )}dx=\int\left (\frac{A}{x-2}+\frac{B}{x-3} +\frac{C}{x-4}\right)dx

Using partial fractions,

x-1=A\left ( x-3 \right )\left ( x-4 \right )+B\left ( x-2 \right )\left ( x-4 \right )+C\left ( x-2 \right )\left ( x-3 \right )

Put  x=3,

B=-2

Put  x=2,

A=\frac{1}{2}

Put x=4,

c=\frac{3}{2}

Now,

\int\frac{x-1}{\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )}dx=\int\left (\frac{\frac{1}{2}}{x-2}+\frac{-2}{x-3} +\frac{\frac{3}{2}}{x-4}\right)dx

\int\frac{x-1}{\left ( x-2 \right )\left ( x-3 \right )\left ( x-4 \right )}dx=\frac{1}{2}log\left | x-1 \right |-2log\left | x-3 \right |+\frac{3}{2}log\left | x-4 \right |+C

Posted by

Suraj Bhandari

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