Unwanted constants after integration

1. Mar 30, 2009

Cyosis

While integrating a rational function I stumbled upon the following problem (In the calculation of the integral the substitution $$u=x+1, du=d(x+1)=dx$$ was used).

\begin{align} \int \frac{x^2}{(1+x)^2}\,dx &= \int \frac{(u-1)^2}{u^2}\,du \\ &= \int du+\int \frac{du}{u^2}\ -2 \int \frac{du}{u}\ \\ &= u-\frac{1}{u}- 2 \log(u) \\ &=1+x - \frac{1}{1+x}-2 \log(1+x) \end{align}

The problem now is that if I substitute x back into the integral during step (2) I get $$x - \frac{1}{1+x}-2 \log(1+x)$$.

Obviously taking the derivative of both primitives yields the same integrand.

My problem with this is that instead of getting an unknown constant I get this unwanted extra 1. Secondly if I plug this integral into mathematica it gives the result without the constant 1.

So my question is why do I get different functions without having specified the integration constant?

2. Mar 30, 2009

CompuChip

You didn't specify the integration boundaries, so you will get an integration constant. So instead of (3), you get u - 1/u - 2 log(u) + C and instead of (4) you get 1 + x - 1/(1 + x) - 2 log(1 + x) + C.

The extra 1 you have can simply be absorbed in C, i.e. define C' = C - 1 or something like that.

Once you specify the integration boundaries (x from a to b) the extra constant term will be absorbed by the change of boundaries in the substitution (i.e. the u-integral will run from a + 1 to b + 1).

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