Find

$1$x$2$

x$2+64$

$$dx

$.$

Solution

Let

x $=8$ tan$(),$

$2$

$<<$

$2$

$.$

Then

dx $=$

$8$sec$2($$)$

$$d$$

and

x$2+64$

$=$

$64($tan$2()+1)$

$=$

$64$ sec$2()$

$=8|$sec$()|=8$ sec$().$

Thus, we have

dxx$2$

x$2+64$

$=$

$$d$64$ tan$2()8$ sec$()$

$$

$=$

$164$

tan$2()$

$$d$.$