Question

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

Answer 1

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

\(Let\  2x=t ⇒ 2dx=dt\)

\(When \ x = 1, t = 2 \ and\  when \ x = 2, t = 4\)

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

\(Let \ \frac 1t=ƒ(t)\)

\(Then,\ ƒ(t)=-\frac {1}{t^2}\)

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

= \([e^tƒ(t)]_2^4\)

= \([e^t.\frac 2t]_2^4\)

= \([\frac {e^t}{t}]_2^4\)

= \(\frac {e^4}{4}-\frac {e^2}{2}\)

= \(\frac {e^2(e^2-2)}{4}\)