Solve $\int \sqrt{7x + 4}\,dx$

Question

I need to solve the following integral

$$\int \sqrt{7x + 4}\,dx$$

I did the following steps:

\begin{align} \text{Let} \, u &= 7x+4 \quad \text{Let} \, du = 7 \, dx \\ \int &\sqrt{u} \, du\\ &\frac{2 (7x+4)^{3/2}}{3} \end{align}

The solution is: $\frac{2 (7x+4)^{3/2}}{21}$. I am having some trouble understanding where the denominator, $21$, comes from (is it because you integrate $du$ also and thus, $7 dx$ becomes $\frac{1}{7}$?). I believe this is some elementary step that I am missing. Can someone please explain to me this?

Thanks!

P.S Is it correct to say "solve the integral"?

Answer 1

I suspect that you mean "calculate the integral $\int\sqrt{7x+4}\,dx.$" Your method was fine, except you have $du=7dx,$ so $dx=\frac17du,$ whence $$\int\sqrt{7x+4}\,dx=\int\sqrt{u}\left(\frac17du\right)=\frac17\int\sqrt{u}\,du.$$

Answer 2

When you made the u-substitution, you took $u=7x+4$ and hence $du=7 dx$. You forgot this factor of 7! In particular, $dx=du/7$.

It helps to write out the $dx$ in the integral: $$\int \sqrt{7x+4} dx=\int \frac{\sqrt{u}}{7} du.$$