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#### mathworker

##### Well-known member

- May 31, 2013

- 119

This is the integral I am trying to evaluate. I would very much appreciate any help. \[\int \frac{2x^3-1}{x+x^4}dx\]

\[\int \frac{2x^3-1}{x+x^4}dx\]

\[\int \frac{1}{2}.(\frac{4x^3+1}{x^4+x}-\frac{3}{x^4+x})dx\]

\[\frac{1}{2}\log{x+x^4}-\frac{1}{2}\int \frac{3}{x^4+x})dx\]

\[\int \frac{3}{x^4+x}dx\]

\[t=\log{x}\]

\[dt=\frac{dx}{x}\]

\[\int \frac{3}{x^4+x}dx=\int \frac{3}{1+e^{3t}}dt\]

**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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