Question

integral of dx/x^3 + 4x from 1 to 2

Answer 1

looks like partial fractions to me.

Answer 2

integration by parts?

Answer 3

i would use partial fractions. in other words write
\[\frac{1}{x(x^2+4)}=\frac{1}{4x}-\frac{x}{4(x^2+4)}\]

Answer 4

then the integration is easy the hard part is figuring out what i wrote. do you know how to do that?

Answer 5

umm no, i do not

Answer 6

have you seen partial fraction decomposition before?

Answer 7

i don't think i've learned that yet, so could you do it by integration by parts?

Answer 8

start by factoring the denominator. so you have
\[\frac{1}{x(x^2+4)}\] then rewrite as
\[\frac{a}{x}+\frac{bx+c}{x^2+4}\] and solve for
\[a,b,c\]

Answer 9

to be honest i am not sure how you would do this by parts, but there is more than one way to skin a cat. maybe amistre or jamesj can help with that part. it is pretty easy once you have broken it into two pieces

Answer 10

okay thank you for trying!

Answer 11

is this from a text, because it really looks like a set up for partial fractions

Answer 12

\[\frac{d}{dx}\ tan^{-1}(\frac{x}{a})=\frac{1}{a}*\frac{1}{\frac{x^2}{a^2}+1}=\frac{1}{\frac{x^2}{a}+a}=\frac{a}{x^2+a^2}\]
maybe?

Answer 13

i am sticking with partial fractions. really i see no other way, and i have a feeling that although there might be one it will be extremely messy. once you have the partial fractions the integral pops right out. it is
\[\frac{1}{4}\ln(x)-\frac{1}{8}\ln(x^2+4)\]

Answer 14

partials are nice :)

Answer 15

that is what my dentist tried to tell me

Answer 16

\[\int\frac{1}{2x}*\frac{2}{(x^2+2^2)}dx\]
might work for IBP

Answer 17

or im over thinking it :)

Answer 18

\[\int\frac{1}{x(x^2+2^2)}dx=\frac{ln(x)}{(x^2+2^2)}-\int ln(x)..dx\]lol