Question

((x+4)\(x^(2)+2x+5))dx

Answer 1

If we complete the square in the denominator we get something like this:
\[\Large\bf\sf \int\limits \frac{x+4}{x^2+2x+5}dx\quad=\quad \int\limits \frac{(x+1)+3}{(x+1)^2+4}dx\]
From here we'll want to apply a Trig Substitution.

Answer 2

It looks like the substitution we want to make is:\[\Large\bf\sf x+1\quad=\quad 2 \tan \theta\]
Any confusion on how I came up with that?

Answer 3

yes...I thought there would be a tan^-1x ( also known as tan inverse) because of 1/1+x^(2)
also I have the answer- I just don't know how to get to it.
the answer is (1\2)ln(x^(2)+2x+5)+(3/2)tan^(-1)*(x+1/2)+C

Answer 4

Yes we'll have an arctan showing up in the solution.
Ok let's go a little further and see where it shows up.

Answer 5

\[\Large\bf\sf dx\quad=\quad 2\sec^2\theta \;d \theta\]

Answer 6

Plugging in our substitution:\[\Large\bf\sf \int\limits \frac{(2\tan \theta)+3}{4\tan^2\theta+4}2\sec^2\theta \;d \theta\]

Answer 7

After simplifying things down a bit (assuming I didn't make a mistake), we should get:\[\Large\bf\sf \frac{1}{2}\int\limits 2 \tan \theta+3\;d \theta\]

Answer 8

So integrating gives us:\[\Large\bf\sf -\ln|\cos \theta|+\frac{3}{2}\theta+C\]

Answer 9

Undoing our substitution is a little tricky.
It requires that we go back to triangle trig.

Answer 10

\[\Large\bf\sf 2\tan\theta\quad=\quad x+1\qquad\to\qquad \tan\theta\quad=\quad \frac{x+1}{2}\]Recall that tan is opposite or adjacent, so we'll draw this relationship.

Answer 11

Get confused along the way anywhere?
I know it's a lot of stuffs so far :o

Answer 12

the only part I got confused was at the x+1=2tan(angle) where did the 2 come from in front of the tan? this is very helpful though!!!(:

Answer 13

So our denominator was:\[\Large\bf\sf (x+1)^2+4\]
We want to make a trig sub and turn it into something like:\[\Large \tan^2\theta+1\]Because that allows us to get rid of the subtraction between them using our trig identity.
See how the 4 is causing a problem though?
When we attach a 2 to our tangent, (and square it), it deals with the 4 nicely.\[\Large\bf\sf (2\tan \theta)^2+4\quad=\quad 4\tan^2\theta+4\quad=\quad 4(\tan^2\theta+1)\]
\[\Large\bf\sf =\quad 4(\sec^2\theta)\]

Answer 14

allows us to get rid of the addition* blah typo.

Answer 15

ohhhh...OK, that makes sense

Answer 16

To finish this up we only have a couple more steps from here.
We need to find the hypotenuse of that triangle (using the Pythagorean Theorem).

Answer 17

Ok so we get something like that, yes?