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?


  • $\begingroup$ Why do "you think" you're totally wrong? Differentiate what you got and check whether you get the original function in the integral! $\endgroup$
    – Timbuc
    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:

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.