∫dx(x 1)(x 2) #10; #10; put x 1=t #10; #10; dx=dt #10; #10; ∫dtt(t 1) #10; #10; =∫ #10;t^12(1 1t dt ) #10; #10; ...

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Integration by Substitution

∫1/x - 1x - 2...

Question

$\int \frac{1}{(x - 1) (x - 2)} d x$

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Solution

$\int \frac{d x}{(x - 1) (x - 2)}$

$p u t x - 1 = t$

$d x = d t$

$\int \frac{d t}{t (t - 1)}$

$= \int t^{\frac{1}{2}} (1 - \frac{1}{t} d t)$

$p u t 1 - \frac{1}{t} = u$

$\frac{1}{t^{2}} d t = d u$

$= \int \frac{d u}{4}$

$= log u + c$

$= log (1 - \frac{1}{x - 1}) + c$

$= log (\frac{x - 2}{x - 1}) + c$

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