How to solve $\int\frac{2x}{x^2-2x+1}dx$ using improper fractions

Question

How would one solve $\int\frac{2x}{x^2-2x+1}dx$ using improper fractions? I tried setting up the integral as equal to $\frac{A}{x-1}+\frac{B}{(x-1)^2}$, but I'm not getting any answer and I don't know what to do. Any assistance you could give would be greatly appreciated.

Answer 1

Instead of resorting to partial fractions, try this

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

which can then be re-written as $$\int \frac {2x-2}{x^2-2x+1}dx+\int \frac {2}{x^2-2x+1}dx.$$

The first part becomes $$\ln(x^2-2x+1)$$ while the second part is $$ 2 \int \frac{1}{(x-1)^2}dx=-2(x-1)^{-1}.$$

So the final answer is simply $$\ln(x^2-2x+1)-2(x-1)^{-1}+C.$$

Answer 2

You can solve it using Partial fractions and get the correct answer. Indeed $$\int \frac{2x}{( x-1)^{2}} \ =\ \int \frac{A}{( x-1)} +\int \frac{B}{( x-1)^{2}}$$ Solving for $A$ and $B$ by plugging in $A(x-1)+B = 2x$, we get $x(A-2) + B-A = 0$, thus $A=B=2$. Thus $$\int \frac{2}{( x-1)} +\int \frac{2}{( x-1)^{2}} \ = 2\ln|x-1| -\frac{2}{( x-1)}+C$$

Answer 3

This is a simpler case of partial fraction decomposition. This can be done without setting $\frac{2x}{x^2-2x+1} = \frac{A}{x-1} + \frac{B}{(x-1)^2}$. I'll show you.

$$\begin{align} \int \frac{2x}{x^2 - 2x + 1}\ dx &= 2\int \frac{x}{(x-1)^2}dx\\& = 2 \int \frac{x- 1 + 1}{(x-1)^2}\ dx \\& = 2 \int \frac{x-1}{(x-1)^2} + \frac{1}{(x-1)^2}\ dx\\& = 2 \int \frac{1}{x-1} + \frac{1}{(x-1)^2}\ dx\\& =\boxed{ 2 \ln|x-1| - \frac{2}{x-1} + C}\end{align}$$