# Finding the indefinite integral of a root function

I'm stuck on a particular integral problem. The problem is stated as:

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

My working thus far:

$$\int x \sqrt{2x+1} dx = \frac{1}{2}x^2\frac{2}{3}(2x+1)^\frac{3}{2}$$ $$\frac{1}{3}x^2(2x+1)\frac{3}{2}$$

...

I think I am totally wrong here and completely on another course. Will somebody please point out where I'm screwing up, and possibly guide me to the correct answer?

Thanks!

• Why do "you think" you're totally wrong? Differentiate what you got and check whether you get the original function in the integral! Sep 19 '14 at 14:55

It looks like to integrate the product, you found the product of the integration of each factor. We can't do that!

If $\int f(x) \,dx = F(x)$ and $\int g(x)\,dx = G(x)$

$$\int f(x)\cdot g(x) \,dx \neq F(x)G(x) + C$$

Let's start over.

Note that since the integrand is defined only for $2x+1\geq0$, we put

$$\underbrace{u^2 = 2x+1}_{u = \sqrt{2x+1}} \implies 2u\,du = 2\,dx\iff u\,du = dx$$

Note also that we also have $2x = u^2 -1 \iff x = \frac 12( u^2 - 1)$.

Then $$I = \int x \sqrt{2x+1} dx = \frac 12\int (u^2 -1)u\cdot u \,du =\frac 12\int(u^4 - u^2)\,du$$

\begin{align} I & = \frac 12\left( \frac {u^5}{5} - \frac{u^3}{3}\right) + C \\ & = \frac 12 \left(\frac{(2x + 1)^{5/2}}{5} - \frac{(2x+1)^{3/2}}{3}\right) + C\end{align}

The integral of a product is not the product of the integrals. Try to put $\ \sqrt{2x+1}=t$

so you have: $\ 2x+1=t^2 \implies x=\frac{t^2-1}{2}$, $dx=tdt$ and now it should be easier:

$\int{x \sqrt{2x+1}dx }=\int{\frac{t^2-1}{2}t^2dt}=\frac{1}{2}\int(t^4-t^2)dt$

• Thanks, Mosk. If I put the root function equal to t, then, how do I proceed with finding its antiderivative? Sep 19 '14 at 11:40
• now you should get it
– Mosk
Sep 19 '14 at 12:07

Put u=2x+1. Then $\frac{$d$u}{$d$x}=2 \implies$d$u=2$d$x$.

Then, $\int x \sqrt{2x+1}= 0.5 \int 2x \sqrt{2x+1}$d$x=0.5 \int (\frac{u-1}{2}) \sqrt u$ d$u$

Which is easy to integrate.