int_{5}^{10} frac{1}{(x-1)(x-2)} d x is equal | vedclass

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Question

(B) Let $I = \int_{5}^{10} \frac{1}{(x-1)(x-2)} dx$. Using partial fractions,we write $\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$. Solving

Vedclass · Accepted Answer

$\log \frac{32}{27}$

Vedclass · Answer

(B) Let $I = \int_{5}^{10} \frac{1}{(x-1)(x-2)} dx$. Using partial fractions,we write $\frac{1}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}$. Solving for constants,we get $A = -1$ and $B = 1$. Thus,$I = \int_{5}^{10} \left( \frac{-1}{x-1} + \frac{1}{x-2} ight) dx$. Integrating,we get $I = [-\log|x-1| + \log|x-2|]_{5}^{10}$. $I = [\log|\frac{x-2}{x-1}|]_{5}^{10}$. Substituting the limits,$I = \log|\frac{10-2}{10-1}| - \log|\frac{5-2}{5-1}|$. $I = \log(\frac{8}{9}) - \log(\frac{3}{4})$. $I = \log(\frac{8/9}{3/4}) = \log(\frac{8}{9} 	imes \frac{4}{3}) = \log(\frac{32}{27})$.

$\int_{5}^{10} \frac{1}{(x-1)(x-2)} d x$ is equal to

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