So I tried solving this by taking a substitute for the integrand, $t=49x$, so its derivative is $dx = \frac {dt} {49}$. Then you insert it into the integrand and get $$\int \frac{\mathrm dt}{49(t^2 +1)} = \frac{1}{49}\int\frac{1}{t^2 +1}dt = \frac{1}{49}\arctan t + c = \frac{1}{49}\arctan 49x + c $$ 

Why is this not correct? Why do I have to take substitute $t=7x$ and then do the thingy? I don't know what I'm doing incorrect.