Compute $\int \frac{\mathrm{d}x}{49x^2+1}$ 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.

\begin{align}
& \int\frac{dx}{49x^2+1} = \int\frac{dx}{(7x)^2+1} = \frac 1 7\int \frac{7\,dx}{(7x)^2+1} = \frac 1 7 \int\frac {dt}{t^2+1} \\[10pt]
= {} &  \frac 1 7 \arctan t + C = \frac 1 7 \arctan(7x)+C
\end{align}
