How do you use substitution to integrate x(sqrt(2x+1))dx.?

Jun 10, 2015

This is better-solved using integration by parts. There is nothing u-substitution-related that is easy to think of quickly, here.

I would choose the function that is easier to differentiate and make go away as $u$, such as ${x}^{n}$.

Let:

• $u = x$
• $\mathrm{du} = \mathrm{dx}$
• $\mathrm{dv} = {\left(2 x + 1\right)}^{1 / 2} \mathrm{dx}$
• $v = \frac{2 / 3}{2} {\left(2 x + 1\right)}^{3 / 2} = \frac{1}{3} {\left(2 x + 1\right)}^{3 / 2}$

Then, we get:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$= \frac{x}{3} {\left(2 x + 1\right)}^{3 / 2} - \int \frac{1}{3} {\left(2 x + 1\right)}^{3 / 2} \mathrm{dx}$

$= \frac{x}{3} {\left(2 x + 1\right)}^{3 / 2} - \frac{1}{3} \left(\frac{1}{5} {\left(2 x + 1\right)}^{5 / 2}\right)$

$= \frac{x}{3} {\left(2 x + 1\right)}^{3 / 2} - \frac{1}{15} {\left(2 x + 1\right)}^{5 / 2}$

$= {\left(2 x + 1\right)}^{3 / 2} \left[\frac{x}{3} - \frac{1}{15} \left(2 x + 1\right)\right]$

$= \frac{1}{15} {\left(2 x + 1\right)}^{3 / 2} \left[5 x - \left(2 x + 1\right)\right]$

And now we tack on the constant at the end.

$= \textcolor{b l u e}{\underline{\frac{1}{15} {\left(2 x + 1\right)}^{3 / 2} \left(3 x - 1\right) + C}}$

Aug 1, 2017

Explanation:

$\int x \sqrt{2 x + 1} \mathrm{dx}$

Let $u$ be the radicand (the stuff under the radical).

So $u = 2 x + 1$, then $x = \frac{1}{2} \left(u - 1\right)$ and $\mathrm{du} = 2 \mathrm{dx}$

$\int x \sqrt{2 x + 1} \mathrm{dx} = \frac{1}{2} \int x {\left(2 x + 1\right)}^{\frac{1}{2}} 2 \mathrm{dx}$

$= \frac{1}{2} \int \frac{1}{2} \left(u - 1\right) {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{1}{4} \int \left({u}^{\frac{3}{2}} - {u}^{\frac{1}{2}}\right) \mathrm{du}$

$= \frac{1}{4} \left(\frac{2}{5} {u}^{\frac{5}{2}} - \frac{2}{3} {u}^{\frac{3}{2}}\right) + C$

$= \frac{1}{10} {u}^{\frac{5}{2}} - \frac{1}{6} {u}^{\frac{3}{2}} + C$

Simplify as desired before reversing the substitution.

I like

$\frac{\sqrt{u}}{30} u \left(3 u - 5\right) + C$

So we have

$\int x \sqrt{2 x + 1} \mathrm{dx} = \frac{\sqrt{2 x + 1}}{30} \left(2 x + 1\right) \left(3 \left(2 x + 1\right) - 5\right) + C$

$= \frac{\sqrt{2 x + 1}}{30} \left(2 x + 1\right) \left(6 x - 2\right) + C$

$= \frac{\sqrt{2 x + 1}}{15} \left(2 x + 1\right) \left(3 x - 1\right) + C$