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Evaluate the integrals in Exercises 1 to 8 using substitution.

    Q8.    \int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx

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\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx
let 2x =t \Rightarrow 2dx =dt
when x = 1 then t = 2 and when x = 2 then t= 4

\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\
let
 \frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}
\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt
                                          \\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}
                                        

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manish

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