Question

Find the indefinite integral. (Remember to use ln |u| where appropriate.)
(x^3 − 3x^2 + 5x − 12)/(x^2 + 4) dx

Answer 1

its an indefinite integral but im lost on how to even solve it

Answer 2

have you divided?

Answer 3

x-3
-----------------------------
x^2+4 | x^3 -3x^2 +5x -12
-(x^3 +4x)
-----------------------
-3x^2+x-12
-(-3x^2 -12)
--------------
x

Answer 4

so i got
\[\frac{x^3-3x^2+5x-12}{x^2+4}=x-3+\frac{x}{x^2+4}\]
if i did my arithmetic right above

Answer 5

yup yup good arithmetic

Answer 6

you did :)

Answer 7

\[\int\limits_{}^{}(x-3+\frac{x}{x^2+4})dx\]

Answer 8

we can use substitution for the 3rd integral
\[\int\limits_{}^{}x dx+\int\limits_{}^{}-3 dx+\int\limits_{}^{}\frac{x}{x^2+4} dx\]

Answer 9

lol

Answer 10

\[\frac{x^2}{2}-3x+\int\limits\limits_{}^{}\frac{x}{x^2+4}dx \]

Answer 11

\[u=x^2+4 => du=2x dx =>\frac{1}{2} du=x dx\]
\[\frac{x^2}{2}-3x+\frac{1}{2}\int\limits_{}^{} \frac{du}{u}+C=\frac{x^2}{2}-3x+\frac{1}{2}\ln|u|+C\]

Answer 12

\[\frac{x^2}{2}-3x+\frac{1}{2}\ln(x^2+4)+C \]

Answer 13

I like that you dropped the absolute value since you didn't need it in the end :)

Answer 14

:)
right because x^2+4>0

Answer 15

so whats the final answer?

Answer 16

:O

Answer 17

i'm puzzled

Answer 18

did you not follow anything i did?

Answer 19

lol

Answer 20

I would guess no :)

Answer 21

I need answer now!

Answer 22

MIA....

Answer 23

ugh my comp keeps freezing sorry