First slide

Methods of integration

Question

$\int \frac{1}{x^{2}} e^{\frac{x - 1}{x}} d x =$

Easy

A

$e^{\frac{x}{x - 1}} + c$
B

$e^{\frac{x - 1}{x}} + c$
C

$- e^{\frac{x}{x - 1}} + c$
D

$- e^{\frac{x - 1}{x}} + c$

Solution

                $1 - \frac{1}{x} = t \Rightarrow \frac{1}{x^{2}} d x = d t$
               $\int \frac{1}{x^{2}} e^{1 - \frac{1}{x}} d x = \int e^{t} d t$
                                   $= e^{t} + c = e^{\frac{x - 1}{x}} + c$

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