Therefore, to find

$1$

$91+$ t$2$

j $+$

$8$t$71+$ t$8$

k

$$dt$,0$

we'll first find

$1$

$91+$ t$2$

$$dt$\mathrm{.}0$

Recall that

$11+$ t$2$

is the derivative of tan$1$t$.$ Therefore, we have

$1$

$91+$ t$2$

$$dt$0$

$=$

tan$1($t$)$

$10$

$=$

$4$

$.$