Question

∫ (x+4)/(x^2 +2x+5) dx

Answer 1

Can you factor the bottom at all?

Answer 2

no, it has no real zeros

Answer 3

so would you have to divide x+4 by the denominator?

Answer 4

i would complete the square on the bottom, then use a trig sub, that might work

Answer 5

you cant use partial fractions because the denominator has not real roots, unless you want to be fancy and use partial fractions with complex numbers
i would not go that route

Answer 6

yeah thats where i got stuck...

Answer 7

denominator can be written as \((x+1)^2+4\) which looks very much like \(u^2+a^2\) with \(u=x+1,b=2\)

Answer 8

ok so, let's make that bottom into something we can use

Answer 9

set up for arctangent

Answer 10

ok can you go through how you got to arctangent...

Answer 11

So, take a look at the bottom. We want to factor that into something we can use. Start by completing the square

Answer 12

\[\int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{u}{a})\]

Answer 13

break in to two parts
use a u-sub for the first part, and arctangent for the second

Answer 14

@FibonacciChick666 you can message me now
i think it is ok once you use a gimmick, as all of these require a gimmick of some sort

Answer 15

if it was
\[\int\frac{x+1}{x^2+2x+5}dx\]you could use a simple u - sub \(u=x^2+2x+5, du =(2x+2)dx,\frac{1}{2}du=(x+1)dx\) etc then it would be easy

Answer 16

since it is not \[\int\frac{x+1}{x^2+2x+5}dx\] but rather \[\int\frac{x+4}{x^2+2x+5}dx\] gimmick is to break in to two parts
\[\int\frac{x+1}{x^2+2x+5}dx+\int\frac{3}{x^2+2x+5}dx\]

Answer 17

first is the simple u - sub as above, you can pretty much do it in your head
second is the complete the square get the arctangent part

Answer 18

ok so after you manipulate it with the trig sub, would you eventually get to something like 1/2∫x+4 ?

Answer 19

not quite.

Answer 20

ok where did I go wrong, i had the denominator being 4(sec^x), and that canceled out with the du = 2 (secx)^2.

Answer 21

It's not a substitution in the form that you are thinking. it is just making it look like something we know how to integrate

Answer 22

ok so how would that look like...

Answer 23

Alright let's walk through step by step. Now, as noted by satelite above, the bottom turns into a pretty \((x+1)^2+4\)

Answer 24

We want to make the top of our fraction something we can use.

Answer 25

So notice that when we make our denominator pretty, it almost looks like a special case we know

Answer 26

Do you see what I'm referring to?

Answer 27

ummm maybe.... i know that u^2 +a^2 allows you to use trig sub, but apparently that was wrong...

Answer 28

ok, so what is the derivative of \(\frac{1}{a}tan^{-1}(\frac{u}{a})\)

Answer 29

1/1+a^2 ?

Answer 30

Why 1?

Answer 31

would it be u?

Answer 32

Can you tell me why? or cite your source?

Answer 33

i just remember that the derivative of arctan is 1/1+x^2

Answer 34

That would be inverse tan of x, I'm asking inverse tan of x/a

Answer 35

1/((x/a)^2 +1) ?

Answer 36

Take a look up top at satelite's answer wayyyy up there

Answer 37

so a u sub for the 1/u, and then the derivative of arctan(u/a) ?

Answer 38

can you write that out for me.

Answer 39

well l/u is the ln|u| right?

Answer 40

no, you're thinking too far ahead

Answer 41

just rewrite, the integral for me

Answer 42

∫(x+4)/(u^2 +4)

Answer 43

ok, so not quite. You are going to have \[\int \frac{x+4}{(x+1)^2+4} dx\] Dp you see how I got that?