Question

Evaluate the integral: $∫_1^2(\frac 1x-\frac {1}{2x^2})e^{2x}\ dx$

Answer 1

$∫_1^2(\frac 1x-\frac {1}{2x^2})e^{2x}\ dx$

$Let\ 2x=t ⇒ 2dx=dt$

$When \ x = 1, t = 2 \ and\ when \ x = 2, t = 4$

∴$∫_1^2(\frac 1x-\frac {1}{2x^2})e^{2x}\ dx$ = $\frac 12∫_2^4(\frac 2t-\frac {2}{t^2})e^t\ dt$

$Let \ \frac 1t=ƒ(t)$

$Then,\ ƒ(t)=-\frac {1}{t^2}$

⇒$$∫_2^4(\frac 1t-\frac {1}{t^2})e^t\ dt = $∫_2^24e^t[ƒ(t)+ƒ'(t)]dt$

= $[e^tƒ(t)]_2^4$

= $[e^t.\frac 2t]_2^4$

= $[\frac {e^t}{t}]_2^4$

= $\frac {e^4}{4}-\frac {e^2}{2}$

= $\frac {e^2(e^2-2)}{4}$