Question

integrate[1/(x^3-5x^2)]

Answer 1

\[\int\limits_{}^{}\frac{1}{x^2(x-5)} dx\]

Answer 2

use partial fractions! :)

Answer 3

\[\frac{1}{x^2(x-5)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-5}\]

Answer 4

we have :1/(x^3 - 5x^2) = A/x + B/x^2 + C/(x - 5),

Answer 5

\[1=Ax(x-5)+B(x-5)+Cx^2\]

Answer 6

myin will do the rest :)

Answer 7

\[1=Ax^2-5Ax+Bx-5B+Cx^2=x^2(A+C)+x(-5A+B)+(-5B)\]

Answer 8

\[A+C=0, -5A+B=0, -5B=1\]

Answer 9

\[-5B=1=> B=\frac{-1}{5}\]

Answer 10

\[-5A+B=-5A+\frac{-1}{5}=0 =>A=\frac{-1}{25}\]

Answer 11

\[A+C=\frac{-1}{25}+C=0 => C=\frac{1}{25}\]

Answer 12

so we have
\[\int\limits_{}^{}(\frac{-1}{25(x)}+\frac{-1}{5x^2}+\frac{1}{25(x-5)}) dx \]

Answer 13

=\[-\frac{1}{25}\ln|x|-\frac{1}{5}\int\limits_{}^{} x^{-2} dx+\frac{1}{25} \ln|x-5|+C\]

Answer 14

\[-\frac{1}{25}\ln|x|-\frac{1}{5} \cdot \frac{x^{-2+1}}{-2+1}+\frac{1}{25} \ln|x-5|+C\]

Answer 15

\[-\frac{1}{25}\ln|x|-\frac{1}{5} \cdot \frac{x^{-1}}{-1}+\frac{1}{25} \ln|x-5|+C\]

Answer 16

\[-\frac{1}{25}\ln|x|-\frac{1}{5} \cdot \frac{-1}{x}+\frac{1}{25} \ln|x-5|+C\]

Answer 17

\[-\frac{1}{25}\ln|x|+\frac{1}{5x}+\frac{1}{25} \ln|x-5|+C\]

Answer 18

mon amie, i like the way you do math

Answer 19

lol

Answer 20

Write a math book

Answer 21

Optionally, you can teach me everything you know

Answer 22

that's not a lot
zarkon has more knowledge

Answer 23

It's not about that. Lagrange is right. Your approaches are consistently brilliant

Answer 24

lol i would hope so since i'm a computer

Answer 25

You're a what? A computer? Oh dear....

Answer 26

lol no i'm kidding

Answer 27

im sure more that just her approach is brilliant

Answer 28

I agree