Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $

Question

Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $

I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.

In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx $

but I am not sure how this was accomplished. Any ideas? I am more interested in the method than the answer.

Answer 1

Since $x^2+x-2 = (x-1)(x+2)$, we can use partial fraction decomposition:

$$\frac{x-2}{x^2+x-2} = \frac{A}{x-1} + \frac{B}{x+2}.$$

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

Equating the coefficients of each power of $x$, we have that $1=A+B$, and $-2=2A-B$.

Solving for $A$ and $B$ decomposes our original rational expression into two simple fractions.

Answer 2

Hint: you can divide the numerator with the denominator and rewrite the integral in this way: $$\int q(x)+\int D(x)/R(x)$$

Answer 3

when you get the $\int 1 - \frac{x-2}{x^2+x-2} dx$, you can use $x^2+x-2 = (x-1)(x+2)$. then it become $$ \int 1 - \frac{x+2-4}{(x+2)(x-1)}dx $$ which equal to $$ \int 1 - \frac{1}{x-1}-\frac{4}{(x+2)(x-1)}dx $$ $$ \int 1 - \frac{1}{x-1}-4/3(\frac{1}{(x-1)}+\frac{1}{(x+2)})dx $$

Answer 4

While partial fraction expansion is certainly the natural choice, here is a slightly different approach that continues from the OP's attempt.

$$\begin{align} \frac{x^2}{x^2+x-2}&=1-\frac{x-2}{x^2+x-2}\\\\ &=1-\frac12 \frac{2x-4}{x^2+x-2}\\\\ &=1-\frac12 \frac{2x+1-5}{x^2+x-2}\\\\ &1-\frac12 \frac{2x+1}{x^2+x-2}+ \frac{5/2}{x^2+x-2}\\\\ &=1-\frac12 \frac{2x+1}{x^2+x-2}+ \frac{5/2}{(x+1/2)^2-(3/2)^2} \end{align}$$

Now, the first term is trivial to integrate. The second term is a perfect differential since the numerator is the derivative of the denominator. And the last term is set up for a hyperbolic trigonometric substitution.