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Question

Mathematics Question on integral

∫xdx(x−1)(x−2) equals∫\frac {xdx}{(x-1)(x-2)}\ equals

A

log∣(x−1)2x−2∣+Clog|\frac {(x-1)^2}{x-2}|+C

B

log∣(x−2)2x−1∣+Clog|\frac {(x-2)^2}{x-1}|+C

C

log∣(x−1x−2)2∣+Clog|(\frac {x-1}{x-2})^2|+C

D

log∣(x−1)(x−2)∣+Clog|{(x-1)(x-2)}|+C

Answer

log∣(x−2)2x−1∣+Clog|\frac {(x-2)^2}{x-1}|+C

Explanation

Solution

Let xdx(x−1)(x−2)\frac {xdx}{(x-1)(x-2)} = A(x−1)+B(x−2)\frac {A}{(x-1)}+\frac {B}{(x-2)}

x = A(x−2)+B(x−1)\frac {A}{(x-2)}+\frac {B}{(x-1)} ...(1)

Substituting x = 1 and 2 in (1), we obtain

A=−1 and B=2A = −1\ and\ B = 2

∴ x(x−1)(x−2)\frac {x}{(x-1)(x-2)} = −1(x−1)+2(x−2)\frac {-1}{(x-1)}+\frac {2}{(x-2)}

⇒ ∫$$\frac {x}{(x-1)(x-2)}dx = ∫$$[\frac {-1}{(x-1)}+\frac {2}{(x-2)}]dx

                             = $-log\ |x-1|+2log\ |x-2|+C$

                             = $log|\frac {(x-2)^2}{x-1}|+C$

Hence, the correct Answer is B