The integral $$\int \frac{2x^3 - 1}{x^4 + x} dx$$, is equal to

We have to evaluate the integral

$$\int \frac{2x^3-1}{x^4+x}\,dx.$$

First, notice that the denominator can be factorised:

$$x^4+x=x(x^3+1).$$

So the integrand becomes

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

Our aim is to split this rational function into simpler parts whose integrals are already known. Because the denominator has a linear factor $$x$$ and an irreducible cubic factor $$x^3+1,$$ we write

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

Multiplying both sides by the common denominator $$x(x^3+1)$$ removes the denominators:

$$2x^3-1=A(x^3+1)+(Bx^2+Cx+D)x.$$

Now we expand and collect like terms.

First expand the right-hand side:

$$A(x^3+1)=Ax^3+A,$$

$$(Bx^2+Cx+D)x=Bx^3+Cx^2+Dx.$$

Adding these gives

$$A x^3 +A + B x^3 + C x^2 + D x.$$

Group powers of $$x$$:

$$x^3\text{ term: }(A+B)x^3,$$

$$x^2\text{ term: }C x^2,$$

$$x\text{ term: }D x,$$

$$\text{constant term: }A.$$

On the left we have $$2x^3-1,$$ which can be rewritten as

$$2x^3+0x^2+0x-1.$$

Hence, equating coefficients of equal powers of $$x$$ on both sides, we obtain the system

$$\begin{aligned} A+B &= 2,\\ C &= 0,\\ D &= 0,\\ A &= -1. \end{aligned}$$

From the last equation $$A=-1.$$ Substituting this value into $$A+B=2$$ gives

$$-1+B=2 \;\Longrightarrow\; B=3.$$

We already have $$C=0$$ and $$D=0.$$ Thus

$$A=-1,\;B=3,\;C=0,\;D=0.$$

Returning to the partial-fraction decomposition, we now know

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

So our integral splits neatly:

$$\int\frac{2x^3-1}{x(x^3+1)}\,dx=\int\left(-\frac{1}{x}\right)dx+\int\frac{3x^2}{x^3+1}\,dx.$$

The first part is immediate. We recall the standard formula

$$\int \frac{1}{x}\,dx=\ln|x|+C.$$

Because we have $$-1/x,$$ we get

$$\int\left(-\frac{1}{x}\right)dx=-\ln|x|.$$

For the second part we perform a simple substitution. We set

$$t=x^3+1.$$

Then

$$\frac{dt}{dx}=3x^2 \quad\Longrightarrow\quad dt=3x^2\,dx.$$

Hence

$$\int\frac{3x^2}{x^3+1}\,dx=\int\frac{dt}{t}=\ln|t|+C=\ln|x^3+1|.$$

Combining both results, we have

$$\int\frac{2x^3-1}{x^4+x}\,dx=\ln|x^3+1|-\ln|x|+C.$$

Using the logarithm property $$\ln a-\ln b=\ln\frac{a}{b},$$ we merge the two logarithms:

$$\ln|x^3+1|-\ln|x|=\ln\left(\frac{|x^3+1|}{|x|}\right).$$

For positive $$x$$ the absolute value on $$x$$ may be dropped, and the final answer can be written succinctly as

$$\log_e\frac{x^3+1}{x}+C.$$

Comparing with the given options, this matches Option C.

Hence, the correct answer is Option C.