Question

This question is to be solved by partial fraction only
integrate((2x^4-x^3-9x^2+x-12)/(x^3-x^2-6x))

Answer 1

i have no idea on how to solve this question please could someone help me.

Answer 2

\[\frac{2x^4-x^3-9x^2+x-12}{x^3-x^2-6x}=2x+1+\frac{4x^2+7x-12}{x^3-x^2-6x}.\][The above is obtained by long division.]
The fraction on the right can be broken down using partial fraction decomposition. To wit, \[\frac{4x^2+7x-12}{x^3-x^2-6x}=\frac{4x^2+7x-12}{x(x-3)(x+2)}=\frac{A}{x}+\frac{B}{x-3}+\frac{C}{x+2}.\]

Answer 3

wow you just made it really simple and easy thanks
and do you know how to solve algebraic substitution if you can explain that to me as well. for example: \[\int\limits(12t-1/\sqrt{4t-3}) dt\]

Answer 4

i tried to solve it but i couldn't do it every time i tried to solve i ended up getting a weird answer could you please explain it to me

Answer 5

Let \(u=4t-3\). Then \(12t-1=12t-9+8=3(4t-3)+8=3u+8.\)Moreover, \(du=4dt\), so \(dt=\frac{du}{4}\). Thus, by substitution, \[\int \frac{12t-1}{\sqrt{4t-3}}dt=\int\frac{3u+8}{\sqrt{u}}\left(\frac{du}{4} \right)=\frac{1}{4}\int(3u^{1/2}+8u^{-1/2})du\]\[=\frac{1}{4}\left(3\frac{u^{3/2}}{3/2}+8\frac{u^{1/2}}{1/2}\right)+C.\]Then just simplify, substitute back \(u=4t-3\) and done. :)

[PS: For the previous question, you must find \(A, B\) and \(C\) first.]

Answer 6

[PS2: After you have found A, B and C, substitute it back, and then you can start integrating.]

Answer 7

ok thank you so much for helping out thou if you don't mind then could you help me solve 1 more question of partial fraction

Answer 8

actually 2 more

Answer 9

one of them i have to prove rule 43 by integration by parts

Answer 10

Sorry @Gurmeet but I have my math class in 15 minutes. :/ Anyway, you can still post the questions here.

Answer 11

ok no problem and thank you so much for helping me i really appreciate it