<s>[INST] $\int \frac {4}{7\sqrt {5x-6}}dx$ 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?
 [/INST] $$\int\frac {4}{7u}\cdot \frac {2u}{5}du = \int \frac{8}{35} du =\frac{8}{35} \color{red}{u} + C = \frac{8}{35} \sqrt{5x-6} + C$$
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