Evaluate the integral.

e$6$x dx

Solution

According to the following statement, we substitute

u $=$

$6$x

$.$

If

nax $+$ b

occurs, we use the rationalizing substitution

u $=$

nax $+$ b

$.$

More generally, this sometimes works for

ng$($x$)$

$.$

Then

$6$x $=$

$,$

so

dx $=$

du

and

e$6$x dx $=$

$13$

ueu du$.$

The integrand is now a product of u and the transcendental function eu so it can be integrated by parts. Choosing

u$=$u $$ dv$=$eu dudu$=$du $$ v$=$

$,$

we have the following.

e$6$x dx

$=$

$13$

$$

eu du

$=$

$$

$13$

$$eu $+$ C

$=$

$$

$13$

$$e$6$x $+$ C$.$