Question

partial fraction

Answer 1

the picture with only one line shows my logic behind the factoring

Answer 2

hold on, i think i'll try another method where the partial fraction is Ax+B/x^2-6x+13

Answer 3

I think that method also is a dead end, i'm unsure what to do but i suspect maybe i need to get (A/x-2)+(B/x-3)+(C/7) as my partial based on the method i used in the pictures

Answer 4

I think you can "complete the square" on the x terms, and get
something that integrates to inverse tangent

Answer 5

\[I=\int\limits \frac{ 1 }{x ^{2}+6x+13 }dx=\int\limits \frac{ 1 }{x ^{2}+6x+9-9+13 }dx\]
\[=\int\limits \frac{ 1 }{\left( x+3 \right)^{2}+2^{2} }dx\]
\[either use \int\limits \frac{ 1 }{x ^{2}+a ^{2} }dx=\frac{ 1 }{ a }\tan^{-1} \frac{ x }{ a }\]
or put x+3=2tan theta ,find dx ,substitute integrate and solve.

Answer 6

correction write -6x then (x-3)^2
put x-3=2tan theta.

Answer 7

i'm trying it on my own quickly and checking with your answer after

Answer 8

I knew I had to complete the square but the method I used (in the pictures) was waaaay off, I got your answer now. thanks for the help!