$\int \frac {4}{7\sqrt {5x-6}}dx$

Question

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?

Answer 1

Note that $$\int dx=x+K$$ So we have that:

$$I=\int\frac {4}{7u}\cdot \frac {2u}{5}du=\frac 8{35}u+K$$ $$I=\frac 8{35}\sqrt {5x-6}+K$$

Answer 2

$$\int\frac {4}{7u}\cdot \frac {2u}{5}du = \int \frac{8}{35} du =\frac{8}{35} \color{red}{u} + C = \frac{8}{35} \sqrt{5x-6} + C$$

Answer 3

I would Substitute $$t=5x-6$$ then is $$x=\frac{t+6}{5}$$ and $$dx=\frac{1}{5}dt$$ The right result is given by $$\frac{8}{35} \sqrt{5 x-6}+C$$

Answer 4

$$\int {4\over 7\sqrt{5x-6}}\mathrm dx\tag1$$

Let $u=5x-6$ then ${du\over dx}=4$, write as $dx={1\over 5}du$

Rewrite $(1)$ in term of $u$

$${4\over 35}\int {\mathrm du\over \sqrt{u}}={4\over 35}\int u^{-1/2}\mathrm du={4\over 35}{u^{-1/2+1}\over -1/2+1}+c={8\over 35}u^{1/2}+c\tag2$$

Rewrite the $(2)$ in term of $x$

$$\int {4\over 7\sqrt{5x-6}}\mathrm dx={8\over 35}\sqrt{5x+6}+c\tag3$$