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.
 A: $49x^2 = (7x)^2$ which is why you should let $t = 7x$. 
A: If you set $t=49x$ then $t^2=49^2x^2$, not $t^2=49x^2$.
A: \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}
A: $$\begin{align}
F(x) &= \int \left(\frac{1}{49x^2 + 1} \right)dx \\
     &= \int \left(\frac{1}{(7x)^2 + 1} \right)dx \\
  t &= 7x \\
  \left(7x \right)dx &= dt \\
  \left(\frac{d}{dx}(7x) \right)dx &= dt \\
  \left(7*\frac{d}{dx}(x) \right)dx &= dt \\
  \left(7*\frac{dx}{dx} \right)dx &= dt \\
  \left(7 \right)dx &= dt \\
  dx &= \left(\frac{1}{7} \right)dt \\
F(t) &= \int \left( \frac{1}{t^2 + 1}*\frac{1}{7} \right)dt \\
     &= \frac{1}{7}*\int \left( \frac{1}{t^2 + 1} \right)dt \\
     &= \frac{1}{7}*\arctan(t) \\
  t &= 7x \\
F(x) &= \frac{1}{7}*\arctan(7x) \\
F(x) &= \bf \left[ \frac{1}{7}*\arctan(7x) + C \right] \\
\end{align}$$
A: Note that $u = 7x \implies u^2 = (7x)^2 =49x^2$
$$u = 7x\implies du = 7dx  \implies$$ $$\int\left(\frac 1{49x^2 + 1}\right)\,dx = \int \frac 17 \left(\frac {7\,dx}{(7x)^2 +1}\right) = \frac 17\int\left(\frac {du}{u^2 + 1}\right)$$
Do you recognize this, now, in the form where you can directly integrate to get $$\frac 17 \arctan(u) + C = \frac 17 \arctan(7x) + C$$
