Evaluate the indefinite integral.

tan$1(6$x$)1+36$x$2$

$$dx

Step $1$

We must decide what to choose for u$.$

If

u $=$ f$($x$),$

then

du $=$ f$($x$)$ dx$,$

so it is helpful to look for some expression in

tan$1(6$x$)1+36$x$2$

$$dx $=$

tan$1(6$x$)$

$11+36$x$2$

$$dx

for which the derivative is also present, though perhaps missing a constant factor.

We see that

tan$1(6$x$)$

is part of this integral, and the derivative of

tan$1(6$x$)$

is

$$$61+(6$x$)2$

$.$

Step $2$

If we choose

u $=$ tan$1(6$x$),$

then

du $=$

$61+36$x$2$

$$dx$.$

If

u $=$ tan$1(6$x$)$

is substituted into

tan$1(6$x$)1+36$x$2$

$$dx$,$

then we have

tan$1(6$x$)1+36$x$2$

$$dx $=$

u$$

$11+36$x$2$

$$dx$.$

We must also convert

$11+36$x$2$

$$dx

into an expression involving u$,$ but we already know that

$11+36$x$2$

$$dx $=$

$$du$.$