Question

How to integrate ((x^2)-3x+7)/((x^2)-4x+6)^2
Please someone help me find the solution..

Answer 1

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x^2-3x+7}{x^2-4x+6}~dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x^2-4x+x+6+1}{x^2-4x+6}~dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x^2-4x+6}{x^2-4x+6}+\frac{x+1}{x^2-4x+6}~dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}1+\frac{x+1}{x^2-4x+6}~dx}\)

Answer 2

This is where I would start, doing a division but that way to make it faster.

Answer 3

The denominator have the power

Answer 4

oh,

Answer 5

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x^2-3x+7}{\left(x^2-4x+6\right)^2}~dx}\)

Answer 6

well, if I do my method I don't know if that is helpful or not you would have:

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(x^2-4x+6\right)}+\frac{x+1}{\left(x^2-4x+6\right)^2}dx}\)

Answer 7

I checked the answer it is rediculous, but it is findable....
What have you attempted so far?

Answer 8

The answer is (7/8)(surd 2)tan inverse((x-2)/surd 2)+((3x-8)/((4x^2)-16x+24)+C

Answer 9

curd?

Answer 10

surd * ?

Answer 11

The wolfram gives this, but that is a long work to get to....
http://www.wolframalpha.com/input/?i=Integral+%28x%5E2-3x%2B7%29%2F%28%28x%5E2-4x%2B6%29%5E2%29

(I assume I am not violating the policy by posting this)

Answer 12

if you start from, then you know integral#1 by completing the square.

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{\left(x^2-4x+6\right)}+\frac{x+1}{\left(x^2-4x+6\right)^2}dx}\)

and integral#2 is what we should figure out.

Answer 13

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x+1}{\left(x^2-4x+6\right)^2}dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x+1}{\left((x+2)^2-10\right)^2}dx}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{(x+2)-1}{\left((x+2)^2-10\right)^2}dx}\)

\(u=x+2\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{u-1}{\left(u^2-10\right)^2}du}\)

Answer 14

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{u-1}{\left(u^2-10\right)^2}du}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{u}{\left(u^2-10\right)^2}-\frac{1}{\left(u^2-10\right)^2}du}\)


u sub and partial fractions for these two respectively

Answer 15

well, r or any other letter sub (not u again)

Answer 16

sorry for overtyping.

Answer 17

Okay..i will try..thanks:)

Answer 18

yw