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An Indefinite Integral

mathworker

Active member
May 31, 2013
118
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:confused:
 
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mathworker

Active member
May 31, 2013
118
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,(Tmi)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
May 31, 2013
118
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
Jan 17, 2013
1,667
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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