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I'm trying to solve this definite integral $$\int_{1}^{e} \frac{x^2-3x^3+1}{2x^3}$$ but the result do not coincide with the one on my book and even in wolframalpha the result is different but I do not see any error in it.

Here my steps:

$$\int_{1}^{e} \frac{x^2-3x^3+1}{2x^3} = \frac{1}{2}\int_{1}^{e} \frac{x^2-3x^3+1}{x^3}$$

$$ \frac{1}{2} \big[ \int\frac{x^2}{x^3}-3\int\frac{x^3}{x^3}+\int\frac{1}{x^3}\big]$$

$$ \frac{1}{2} \big[ ln|x| -3x - \frac{2}{x^2} \big]$$ $$ \big[ \frac{1}{2}ln|x| -\frac{3}{2}x - \frac{1}{x^2} \big]_{1}^{e}$$

And I get:

$$ \frac{ln|e|}{2} - \frac{3}{2}e - \frac{1}{e^2} - \frac{1}{2}ln|1| + \frac{3}{2} + 1 = 3-\frac{3}{2}e - \frac{1}{e^2}$$

So: $$\int_{1}^{e} \frac{x^2-3x^3+1}{2x^3} = 3-\frac{3}{2}e - \frac{1}{e^2}$$

Christian Giupponi
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  • Almost right: observe the third integral: $;\int\frac1{x^3}=-\frac1{2x^2};$ ... – Timbuc Jun 17 '15 at 14:13
  • Sorry but I do not understand where $-\frac{1}{2x^2}$ comes from, in my book I have the rule: $\int \frac{1}{x^n} = - \frac{n-1}{x^(n-1)}$ – Christian Giupponi Jun 17 '15 at 14:23
  • Sorry but I don't understand where your doubt comes from: just differentiate carefully the right side and observe you get the left side! The rule you say your book has is wrong as you can easily check by differentiating. – Timbuc Jun 17 '15 at 14:27
  • Damn, thanks! it works only if n = -1. The rule is $\frac{x^(n+1)}{n+1}$ – Christian Giupponi Jun 17 '15 at 14:31

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