Question

integrate (1/(x^2+2x+2)^2) dx

Answer 1

(x^2+2x+2+1-1)^2=((x+1)^2+1)^2
if you let u=x+1 you get 1/(u^2+1)^2

you should get tan^-1(u^-3)

so your answer should be \[(1/-3)\tan^{-1}( (x+1)^-3)\]

Answer 2

the back of the book is showing a different answer...

Answer 3

whats it showing i didnt do this on scratch paper. does it involve tan_1 or an ln

Answer 4

\[1/2(\tan^{-1} (x+1))+ 1/2((x+1)/(x^2+2x+2))+c\]

Answer 5

complete the square of the denominator -
\[x^2 + 2x + 2 = (x+1)^2 +1\]
Then let
\[x+1 = \tan(y)\]
\[dx = \sec^2(y)dy\]
back sub to find
\[\int \frac{1}{(\tan^2(y)+1)^2}\sec^2(y) dy\]
Simplify to get
\[\int \frac{1}{sec^2(y)}dy\]
Simplify some more
\[\int \cos^2(y)\; dy\]
Use power reduction
\[\frac{1}{2}\int (1-\cos(2y)) dy\]

Answer 6

i just punch this in a derivitive calc and it spit out some thing different than both of ours,

Answer 7

hes right creative thinking right there i forgot to do the d(tanu)

Answer 8

Integrating you get
\[\frac{1}{2}y - \frac{1}{4}\sin(2y) +c\]
probably want to write that as
\[\frac{1}{2}y - \frac{1}{2}\sin(y)\cos(y) +c\]
Now back to the x's, got to set up the triangle.

Answer 9

\[\text{Since }\;\;x + 1 = \tan(y), \text{ then }\;\; y = \tan^{-1}(x+1)\]
And
\[\sin(y) = \frac{x+1}{\sqrt{(x+1)^2+1}}\]
\[\cos(y) = \frac{1}{\sqrt{(x+1)^2+1}}\]

Answer 10

In total, the answer looks like what avenged56 wrote above.

Answer 11

yes i would agree

Answer 12

ill follow through with it and figure out whats going on from there. thanks for the help