Question

\[\int\limits_{1}^{\infty}1/(x ^{2}+4x+3)dx\]

Answer 1

Partial fractions?

Answer 2

i tried integrating it by partial fraction

Answer 3

but i think it's U substitution

Answer 4

\[\int\limits\limits_{1}^{\infty} \frac {1}{x ^{2}+4x+3}dx \rightarrow \lim_{k \rightarrow \infty } \int\limits_{1}^{k} \frac {1}{2(x+1)} - \frac {1}{2(x+3)} dx\]

Answer 5

factor the denominator to get and do your partial fraction decomposition from there.

Answer 6

\[\lim_{k \rightarrow \infty } \int\limits\limits_{1}^{k} \frac {1}{2(x+1)} - \frac {1}{2(x+3)} dx = \lim_{k \rightarrow \infty} \left[ \frac {1}{2} \ln \left| \frac {x+1}{x+3} \right| \right]_{1}^{k}\]

Answer 7

i tried factoring out the denominator but i didn't get the roots right

Answer 8

And that is .5ln2

Answer 9

how did u get 5ln2?

Answer 10

\[\frac {1}{2} \lim_{k \rightarrow \infty} \left[ \ln \left| \frac {x+1}{x+3} \right| \right]_{1}^{k} = \frac {1}{2} \left[ (\lim_{x \rightarrow \infty} \ln \frac {x+1}{x+3}) - \ln \frac {2}{4} \right] = \frac {1}{2} \left[ \ln 1 + \ln 2 \right] = \frac {1}{2} \ln 2\]

Answer 11

oh ok i got it now. thank you

Answer 12

Your welcome =)