Question

how to integrate
1/(x^2+10*x+6)

Answer 1

partial fractions is my bet.

Answer 2

you want the answer or the method?

Answer 3

just checked with wolfram and they give method as some sort of completing the square, getting arctanh and the converting back, but i would use partial fractions. set the denominator= and solve

Answer 4

you get
\[x^2+10x+6=0\]
\[x=-5+\sqrt{19},x=-5-\sqrt{19}\] so you have
\[\int\frac{dx}{(x-(-5+\sqrt{19})(x-(-5-\sqrt{19})}\]

Answer 5

then solve
\[\frac{A}{x-(-5+\sqrt{19})}+\frac{B}{x-(-5-\sqrt{19})}\] for A and B\]

Answer 6

hmm maybe be easier way but i don't see it. not that bad actually but the calculation stinks

Answer 7

oh no it doesn't it is easy.
\[B=\frac{1}{2\sqrt{19}}\] and
\[A=-\frac{1}{2\sqrt{19}}\]

Answer 8

integrate
\[-\frac{1}{2\sqrt{19}}\int\frac{dx}{x+5-\sqrt{19}}\] get
\[-\frac{1}{2\sqrt{19}}\ln(x+5-\sqrt{19})\]
and similarly
\[\frac{1}{2\sqrt{19}}\int \frac{dx}{x+5+\sqrt{19}}=\frac{1}{2\sqrt{19}}\ln(x+5+\sqrt{19})\]