Evaluate the integral (improper integration)

Question

Evaluate $\int\frac{x^4+x+1}{x^3+x}\,dx$.

I need help doing this problem, I started to do long division but I don't know how to because the coefficients don't match.

Thank you

Answer 1

HINT: write the integrand in the form $$x+{x}^{-1}+{\frac {-2\,x+1}{{x}^{2}+1}}$$

Answer 2

By using partial fractions we can see that ,

$\frac{(x^4+x+1)}{(x^3+x)} = \frac {(1-2 x)}{(x^2+1)}+x+\frac {1}{x}$

So $\int \frac{(x^4+x+1)}{(x^3+x)}dx =\int \frac {(1-2 x)}{(x^2+1)}dx+ \int x dx+ \int \frac {1}{x}dx$

$\Rightarrow \int \frac{(x^4+x+1)}{(x^3+x)}dx = \int \frac {1}{(x^2+1)}dx-\int \frac {2 x}{(x^2+1)}dx+ \int x dx+ \int \frac {1}{x}dx$

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

Answer 3

Since: $$\frac{x^4+x+1}{x^3+x} = x+\frac{-x^2+x+1}{x^3+x} = x-\frac{1}{x}-\frac{2x}{1+x^2}+\frac{1}{1+x^2}$$ the primitive of the LHS is given by: $$ \frac{x^2}{2}-\log(x^3+x)+\arctan(x)+C.$$