Evaluate $\int\frac{x}{x+1}dx$

Question

I've not studied calculus for a couple of months so my skills got a little bit rusty.

I'm trying to calculate $\int\frac{x}{x+1}dx$ using the chain rule. Wolfram Alpha says the result is $x - \ln(x+1)$.

So here's where I'm at:

$$\int\frac{x}{x+1}dx = x \cdot \ln(x+1) + \int{\ln(x+1)}dx$$

And I can't figure out how to attack $\int{\ln(x+1)}dx$.

Actually, I need this for an integral test for convergence (I'm trying to find out whether or not the series $\sum\frac{x}{x+1}$ converges), so proving that $x \cdot \ln(x+1) + \int{\ln(x+1)}dx$ is finite for all $x >= 1$ is would be enough.

Answer 1

$\frac{x}{x+1} = \frac{x+1-1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}$

Or, do a $u$-sub with $u=x+1$.

Handle $\int \ln(x+1) \ dx$ with integration by parts with parts $u=\ln(x+1)$ and $dv=dx$.

And, to nitpick, there should be absolute value bars all over the place unless you can assume $x > -1$.

Answer 2

I'm assuming you want to verify the convergence/divergence of

$$\sum_{x=1}^N\frac{x}{x+1}$$

Instead of getting stuck with that integral, see that intuitively each summand looks like $\frac{1}{x}$, a little bit grater actually, and I suppose you have already seen that $$\sum_{x=1}^N\frac{1}{x}$$ diverges.

Let's write it down: since $x\geqslant 1$ you have that $ \frac{1}{x+1}\leqslant\frac{x}{x+1}\ \forall x\geqslant 1$ so

$$\sum_{x=1}^N\frac{1}{x+1} \leqslant \sum_{x=1}^N\frac{x}{x+1}$$

Since the LHS diverges (it can be written as $\sum_{i=2}^{N+1}\frac{1}{i} $ if you take $i=x+1$ ), the RHS too.