$\displaystyle \int \frac{2x-3}{(x-1)^2}dx = \frac{A}{x-1}+\frac{B}{(x-1)^2}$
A=(-1/3)
B=(8/3)
I don't know how to find the Ax^2...Please Help.
$\displaystyle
\begin{gathered}
\frac{{2x - 3}}
{{\left( {x - 1} \right)^2 }} = \frac{A}
{{x - 1}} + \frac{B}
{{\left( {x - 1} \right)^2 }} \hfill \\
2x - 3 = A\left( {x - 1} \right) + B \Leftrightarrow 2x - 3 = Ax - A + B \hfill \\
\end{gathered}
$
$\displaystyle
\Rightarrow \left. {\underline {\,
\begin{gathered}
A = 2 \hfill \\
- A + B = - 3 \hfill \\
\end{gathered} \,}}\! \right| \Rightarrow B = 1 \Rightarrow \frac{{2x - 3}}
{{\left( {x - 1} \right)^2 }} = \frac{2}
{{x - 1}} - \frac{1}
{{\left( {x - 1} \right)^2 }}
$
But look at this:
$\displaystyle
\int {\frac{{2x - 3}}
{{\left( {x - 1} \right)^2 }}} dx = \int {\frac{{2x - 2}}
{{x^2 - 2x + 1}}dx - \int {\frac{1}
{{\left( {x - 1} \right)^2 }}} } dx
$
To the first integral:
$\displaystyle
\int {\frac{{2x - 2}}
{{x^2 - 2x + 1}}dx} = \int {\frac{{\left( {x^2 - 2x + 1} \right)^\prime }}
{{x^2 - 2x + 1}}dx} = \ln \left( {x^2 - 2x + 1} \right) + k
$
The second integral is direct
What Nacho said to do is probably how I'd go about it (especially because some of the other posts are incorrect). BUT, if you have to use partial fractions, then let:
$\displaystyle \frac{2x-3}{(x-1)^2} = \frac{A}{x-1}+\frac{B}{(x-1)^2}$
Multiplying through by $\displaystyle (x-1)^2$ gives you $\displaystyle A(x-1)+B = Ax-A+B = (A)x + (-A+B) = 2x-3$.
Set $\displaystyle A = 2$ and it follows that $\displaystyle B=-1$.
Thus, $\displaystyle \int \frac{2x-3}{(x-1)^2}\,dx = \int \frac{2}{x-1}\,dx-\int\frac{1}{(x-1)^2}\,dx$.
The first integral is simply $\displaystyle 2ln(x-1) = ln((x-1)^2) = ln(x^2-2x+1)$, which if you note, is the same answer Nacho got using his method.
The second integral is $\displaystyle -\frac{1}{x-1}$.
Thus $\displaystyle ln(x^2-2x+1)-(-\frac{1}{x-1}) = ln(x^2-2x+1)+\frac{1}{x-1}$.
$\displaystyle \int \frac{2x-3}{(x-1)^2}\,dx = ln(x^2-2x+1)+\frac{1}{x-1} + C$
I'm not seeing what you're doing here. I think Nacho's work below this post is correctly done. You conclude in your second post in this thread that the original fraction can be replaced by one different fraction - which intuitively doesn't make sense. I think this is an error.