Integral of $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$

Question

I have this simple integral: $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ and I can't come up with the correct answer.

Here's what I did:

I found the roots for $x^3-x^2+x-1$ so I could do partial fractions:

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

I then found the values for $A$, $B$ and $C$:

$x^2(A+B)+x(-B+C)+A-C=-x^2+2x-3$

$ \left\{ \begin{array}{c} A+B=-1 \\ -B+C=2 \\ A-C=-3 \end{array} \right. $

$A=-1$, $B=0$, $C=2$

So now back again with the integral:

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

But why the correct answer is $2 \arctan(x)-\ln|x-1|+C$? Where does the $arctan$ come from?

Answer 1

Your computation is almost correct, but you made a mistake here.

Note that$\int\dfrac{2}{x^2+1}=2\arctan(x)$, but not $\int\dfrac{2}{x^2+1}\ne2|x^2+1|$

Answer 2

The $\arctan$ comes as a result of the antiderivative

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

You calculated that antiderivative wrong: the antiderivatives of fractions are a little more nuanced than just throwing the denominator in a natural logarithm.

Answer 3

You have calculated the problem correctly but you made a small mistake while taking the integral of $\int\dfrac{2}{x^2+1}$

To solve such integrals, take the constant out and then simplify: $$2\cdot \int \:\frac{1}{x^2+1}dx$$

We know, $$\int \:\frac{1}{x^2+1}dx = \arctan(x) $$

So, $$\int\dfrac{2}{x^2+1} = 2\cdot \arctan \left(x\right)$$

As you progress in Calculus, you have to have these kinds of integrals memorized. See derivatives of inverse trig functions.