How Do You Solve This Complex Indefinite Integral?

[mathworker]

This is the integral I am trying to evaluate. I would very much appreciate any help. \[\int \frac{2x^3-1}{x+x^4}dx\]
MY APPROACH:
\[\int \frac{2x^3-1}{x+x^4}dx\]
\[\int \frac{1}{2}.(\frac{4x^3+1}{x^4+x}-\frac{3}{x^4+x})dx\]
\[\frac{1}{2}\log{x+x^4}-\frac{1}{2}\int \frac{3}{x^4+x})dx\]
now we have to find,
\[\int \frac{3}{x^4+x}dx\]
let,
\[t=\log{x}\]
\[dt=\frac{dx}{x}\]
\[\int \frac{3}{x^4+x}dx=\int \frac{3}{1+e^{3t}}dt\]
i am stuck at this point

 

[mathworker]

Re: an indefinite integral

okay wait i think i got it
$$\int \frac{1}{1+e^{3t}}dt$$
$$\int \frac{1+e^{3t}}{1+e^{3t}}dt-\int \frac{e^{3t}}{1+e^{3t}}dt$$
$$t-\frac{log{1+e^{3t}}}{3}$$
sorry for trouble,(Tmi)

 

[MarkFL]

Re: an indefinite integral

The partial fraction decomposition of the integrand is:

$$\frac{2x-1}{x^2-x+1}-\frac{1}{x}+\frac{1}{x+1}$$

 

[mathworker]

Re: an indefinite integral

The partial fraction decomposition of the integrand is:

$$\frac{2x-1}{x^2-x+1}-\frac{1}{x}+\frac{1}{x+1}$$

may be this is the real thought behind he question as it was asked in a competition

 

[alyafey22]

Re: an indefinite integral

Integrals of the form $$\int \frac {1}{x^p+x}dx$$ by taking $ x^p , p> 0$ as a coomon factor so we get $$\int \frac {\frac {1}{x^p}}{1 + \frac {1}{x^{p-1}}} dx$$ . Now it is easy to use a substitution or realize the numerator is the derivative of the denominator after multiplying by a constant .