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Could you please help me solve this integral?

$$\int\frac{1}{x^2-5x+6}dx$$

I know I have to use partial fraction decomposition but I don't know how.

  • 2
    **Hint:** $\dfrac{1}{x^2-5x+6} = \dfrac{1}{x-3}-\dfrac{1}{x-2}$ – 2017-01-27
  • 1
    I guess by decomposing $\frac{1}{x^2-5x+6}$ in partial fractions. – 2017-01-27
  • 1
    What do you know about partial fraction decomposition ? – 2017-01-27
  • 0
    alternativly one might complete the square in the denominator – 2017-01-27

4 Answers 4

2

Let $\displaystyle\frac{1}{x^2-5x+6}=\frac{1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$. This is equivalent to

$1=A(x-3)+B(x-2)$. Letting $x=2$, we have $1=-A$ 0r $A=-1$, and letting $x=3$, we have $1=B$.\

Hence $$\int\displaystyle\frac{1}{x^2-5x+6}dx=\int\bigg(\frac{1}{x-3}-\frac{1}{x-2}\bigg)dx=\ln|x-3|-\ln|x-2|+c$$

1

$$\frac{1}{x^2-5x+6} = \frac{1}{(x-2)(x-3)} = \frac{A}{x-2}+\frac{B}{x-3}\cdots \cdots (1)$$

So $$\frac{1}{(x-2)(x-3)} = \frac{A(x-3)+B(x-2)}{(x-2)(x-3)}\Rightarrow 1= A(x-3)+B(x-2)$$

Now put $x=2$ and $x=3$ you will get $A$ and $B$ and put into $1$

0

$x^2-5x+6=(x-3)(x-2)$, so $\frac{1}{x^2-5x+6}=\frac{(x-2)-(x-3)}{(x-2)(x-3)}=\frac{1}{x-3}-\frac{1}{x-2}$.

That can be easily integrated to get $ln(x-3)-ln(x-2)=ln(\frac{x-3}{x-2})$.

0

$$\int\frac{1}{x^2-5x+6}dx$$

$$ = \int\frac{1}{x^2-3x-2x+6}dx$$

$$ = \int\frac{1}{x(x-3)-2(x-3)}dx$$

$$ = \int\frac{1}{(x-2)(x-3)}dx$$

$$ = \int\left(\frac{1}{x-3}-\frac{1}{x-2}\right)dx$$

$$ = \log(x-3)-\log(x-2)+\log C$$

$$ = \log\Big(\frac{x-3}{x-2}\Big)+ \log C$$