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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?


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

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You had $dx = \frac{du}{7}$, but you wrote only $du$. – Daniel Fischer Aug 15 '13 at 16:36
Maybe "evaluate the integral" instead. – ja72 Aug 15 '13 at 16:56
Solve the integral sounds okay to me. – Ataraxia Aug 15 '13 at 17:00
up vote 3 down vote accepted

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 $$\int\sqrt{7x+4}\,dx=\int\sqrt{u}\left(\frac17du\right)=\frac17\int\sqrt{u}\,du.$$

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Ahh I see but for another question $\int \frac{\cos x}{\sin^2 x}$. The solution let $u = \sin x$ and $du = \cos x \, dx$ but then they didn't do $\frac{du}{\cos x} = dx$ how come? What's the difference? – Jeel Shah Aug 15 '13 at 16:44
@gekkostate You can still use this method, since: $$ \int \dfrac{\cos x}{\sin^2 x}dx = \int \dfrac{\cos x}{u^2} \cdot \dfrac{du}{\cos x} = \int \dfrac{1}{u^2}du $$ – Adriano Aug 15 '13 at 16:50
Ahh! I see now. Thank you! – Jeel Shah Aug 15 '13 at 16:52

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: $$\int \sqrt{7x+4} dx=\int \frac{\sqrt{u}}{7} du.$$

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