An Indefinite Integral

mathworker

Active member
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

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mathworker

Active member
Re: an indefinite integral

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

MarkFL

Staff member
Re: an indefinite integral

The partial fraction decomposition of the integrand is:

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

mathworker

Active member
Re: an indefinite integral

The partial fraction decomposition of the integrand is:

$$\displaystyle \frac{2x-1}{x^2-x+1}-\frac{1}{x}+\frac{1}{x+1}$$
may be this is the real thought behind he question as it was asked in a competition

ZaidAlyafey

Well-known member
MHB Math Helper
Re: an indefinite integral

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

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