Evaluate the integral.

$65$

$($x $1)($x$2+64)$

dx

Step $1$

The given function

$65$

$($x $1)($x$2+64)$

dx

is a rational function of the form

f$($x$)=$

P$($x$)$

Q$($x$)$

where

P$($x$)=65$

and

Q$($x$)=($x $1)($x$2+64).$

Since the degree of the numerator P$($x$)$ is less than the degree of the denominator Q$($x$),$ we do not need to divide. We also note that we cannot factor the denominator more than it is already factored. We further note that

Q$($x$)$

contains irreducible quadratic factors, none of which are repeated. This allows us to find a partial fraction decomposition of the integrand as follows.

$65$

$($x $1)($x$2+64)$

$=$

A

x $1$

Correct: Your answer is correct.

seenKey $1$

$+$

Bx $+$ C

x$2+64$

Correct: Your answer is correct.

seenKey $64$

Step $2$

Combining the fractions, we get

A

x $1$

$+$

Bx $+$ C

x$2+64$

$=$

$($A $+$ B$)$x$2+($C $$ B$)$x $+$

$$

$($x $1)($x$2+64)$

$.$