Thread: [SOLVED] find integral of partial fraction

1. [SOLVED] find integral of partial fraction

$\int \frac{2x-3}{(x-1)^2}dx = \frac{A}{x-1}+\frac{B}{(x-1)^2}$

A=(-1/3)
B=(8/3)

2. see below post

3. $
\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}
$

$
\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:

$
\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:

$
\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

4. post about previous post, containing error. see above post.

5. contains same error

6. 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:

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

Multiplying through by
$(x-1)^2$ gives you $A(x-1)+B = Ax-A+B = (A)x + (-A+B) = 2x-3$.

Set $A = 2$ and it follows that $B=-1$.

Thus, $\int \frac{2x-3}{(x-1)^2}\,dx = \int \frac{2}{x-1}\,dx-\int\frac{1}{(x-1)^2}\,dx$.

The first integral is simply $2ln(x-1) = ln((x-1)^2) = ln(x^2-2x+1)$, which if you note, is the same answer Nacho got using his method.

The second integral is $-\frac{1}{x-1}$.

Thus $ln(x^2-2x+1)-(-\frac{1}{x-1}) = ln(x^2-2x+1)+\frac{1}{x-1}$.

$\int \frac{2x-3}{(x-1)^2}\,dx = ln(x^2-2x+1)+\frac{1}{x-1} + C$

7. Originally Posted by mollymcf2009

$\frac{A}{x-1} + \frac{B}{(x-1)^2}$

$= A(x-1)^2 + B(x-1)$
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.

8. oops. I completely missed that. Thanks guys.

9. this is what I have when I rearrange it:
$6x=x^2(A+B)+x(2A-2B+C)+(4A-2C)$

I'm having trouble solving for A,B, and C. Wouldn't the x(2A-2B+C) = 6?

10. You probably meant to post that here.

But yes, $2A-2B+C=6$. Also $A+B = 0$ and $4A-2C=0$.

Solve the system of equations to find A, B, and C.