Proof Verification: $\frac{x}{x+1} \leq \ln(x+1)$ for $x > -1$ with equality iff $x=0$
My Attempt:
We know that $\frac{1}{(x+1)^2} \leq \frac{1}{x+1}$, for $x > 0. $At $x=0 $ they are the same.
$$\int\frac{1}{(x+1)^2}dx \leq \int\frac{1}{x+1}dx$$ $$\frac{-1}{x+1} +C_1 \leq \ln(x+1) + C_2$$
We conclude that $C_1 =1, C_2 =0$ f0r the $x=0$ case from the original equality to hold.
$$\frac{-1}{x+1}+\frac{x+1}{x+1}-1 +1 \leq \ln(x+1)$$ $$\frac{x}{x+1} \leq \ln(x+1)$$
Is this approach right?
Your answer is ok. But it is better to integrate $$ \frac{1}{(t+1)^2} \leq \frac{1}{t+1}$$ from $0$ to $x$; namely $$ \int_0^x\frac{1}{(t+1)^2}dt \leq \int_0^x\frac{1}{t+1}dt. $$ Then you will get the desired inequality soon instead of concluding $C_1=C_2=0$.