I'm to fault for not going to lecture and missing the lecture on integrals.
I get the basics and trig. But I'm clueless about dividing and what not.
Help?
2006-11-07 11:57:54 · 6 answers · asked by velmakelly777 1 in Science & Mathematics ➔ Mathematics
How do I get x^-2+x^-5 from the equation???
2006-11-07 12:07:12 · update #1
int1/x^2+int1/x^5
x^-1/-1 +x^-4/-4+C
-1/x-1/4x^4+C
2006-11-07 12:05:14 · answer #1 · answered by raj 7 · 0⤊ 0⤋
86 with the power of 395
2006-11-07 11:59:44 · answer #2 · answered by boys <3 me 2 · 0⤊ 0⤋
Anytime uyou have a single denom, it is easiest to brake it apart.
So you want int of x^-2+x^-5
Int of that would be
[x^(n+1)]/(n+1)+C
So
-1x^(-1)+-4x^(-4)+c
---------------------------
you get x^-2+x^-5 by ..
start by asking yourself this,
what is x^2/x^1
When you divide you subtract exponents.
Thus you get x^1
This can also cary over to this problem.
rewrite it as
x^3/x^5 +1/x^5
So
3-5=-2
x^-2+x^-5
2006-11-07 13:03:29 · answer #3 · answered by heyhelpme41 3 · 0⤊ 0⤋
The function is an incorrect fraction, meaning, numerator is bigger than the denominator. best attitude is: do an prolonged branch then combine, do no longer overlook to function the arbitrary consistent, C.
2016-10-21 11:02:16 · answer #4 · answered by ? 4 · 0⤊ 0⤋
(x^3+1)/x^5=
x^-2+x^-5
Integral[x^-2+x^-5]=
(x^-1)/-1+(x^-4)/-4 +C=
(-1/x)-(1/(4x^4))+C
2006-11-07 12:02:03 · answer #5 · answered by Greg G 5 · 0⤊ 0⤋