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$\int\limits^{2}_{1}\frac{x\, dx}{(x+1)(x+2)}=$ |
$tan^{-1}\frac{32}{27}$ $tan^{-1}\frac{27}{32}$ $log\frac{32}{27}$ $log\frac{27}{32}$ |
$log\frac{32}{27}$ |
The correct answer is option (3) → $\log\frac{32}{27}$ $\int\limits^{2}_{1}\frac{x}{(x+1)(x+2)}dx=\int\limits^{2}_{1}\frac{2(x+1)-(x+2)}{(x+1)(x+2)}dx$ $=\int\limits^{2}_{1}\frac{2}{(x+2)}-\frac{1}{(x+1)}dx=\left|\log\frac{(x+2)^2}{(x+1)}\right|^{2}_{1}$ $=\log\frac{32}{27}$ |
