# 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).