Find$$

$1$x$2$

x$2+4$

$$dx

$.$

SOLUTION$$Let$$

x$=2$tan$(),/2<</2.$

$$Then$$

dx$=$

$$

$$d$$

$$and

x$2+4$

$=$

$4$tan$2()+4$

$=$

$4$sec$2()$

$=2|$sec$()|=2$sec$().$

Thus we have

dxx$2$

x$2+4$

$=$

$$

$$d$4$tan$2()2$sec$()$

$$

$=$

$14$

$$

tan$2()$

$$d$.$

To evaluate this trigonometric integral we put everything in terms of sin$()$ and cos$().$

sec$()$tan$2()$

$=$

$1$

$$

$$

cos$2()$sin$2()$

$$

$=$

$$

sin$2()$

Therefore, making the substitution$$

u$=$ sin$(),$

$$we have

dxx$2$

x$2+4$

$=$

$14$

$$

sin$2()$

$$d$$

$=$

$14$

du

$$

$=$

$14$

$$

$1$u

$+$C

$=$

$$

$14$sin$()$

$+$C

$=$

$$

csc$()4$

$+$C$.$