$\int \frac {4}{7\sqrt {5x-6}}dx$ Why can't I use the substitution
$$u=\sqrt {5x-6}$$
$$u^2=5x-6$$
$$2udu=5dx$$
$$dx=\frac {2u}{5}du$$
$$\int\frac {4}{7u}\cdot \frac {2u}{5}du=\int \frac {8u}{35u}du=\int \frac {8}{35}du=\frac {8}{35}+C$$
The answer is obviously incorrect, but what have I done wrong?

Note that 
$$\int dx=x+K$$
So we have that:
$$I=\int\frac {4}{7u}\cdot \frac {2u}{5}du=\frac 8{35}u+K$$
$$I=\frac 8{35}\sqrt {5x-6}+K$$
