How do you find the integral of #(x^4+x-4) / (x^2+2)#?

Question

William Abdalla · Accepted Answer

First, use polynomial long division to get #(x^4+x-4)/(x^2+2)=x^2-2+x/(x^2+2)#.  Next, integrate this expression, using a substitution #u=x^2+2, du=2x\ dx# to get: #\int(x^4+x-4)/(x^2+2)dx=\int(x^2-2+x/(x^2+2))dx# #=\frac{1}{3}x^3-2x+\frac{1}{2}ln(x^2+2)+C# Note that absolute value signs are not needed in the argument of the logarithm function since #x^2+2>0# for all real #x#.

Emma Aldridge · Answer

undefined To find the integral of $\frac{{x^4 + x - 4}}{{x^2 + 2}}$, you can use polynomial long division to divide $x^4 + x - 4$ by $x^2 + 2$, then integrate each resulting term. The division results in $x^2 - 2$ with a remainder of $x + 2$. So the integral becomes:

$\int \frac{{x^4 + x - 4}}{{x^2 + 2}} dx = \int (x^2 - 2) + \frac{{x + 2}}{{x^2 + 2}} dx$

The integral of $x^2 - 2$ is straightforward: $\frac{{x^3}}{3} - 2x$.

For the fraction $\frac{{x + 2}}{{x^2 + 2}}$, perform partial fraction decomposition:

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

Solve for A and B. Then integrate each term separately.

Once integrated, combine the results to get the final integral.