# How do I calculate $\int (x+1)\sqrt {2x+x^2}\;dx$?

I don't see a clear substitution to calculate that since when

$$u = 2x+x^2,\qquad dx = \frac{du}{2+x},$$

And so far as I've understand $dx$ shouldn't be in function of $x$ in order to calculate the integral.

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No, $dx=du/(2+2x)$. Try that, and then do a little algebra before integrating. – Brian M. Scott Sep 27 '11 at 0:24
Better yet, learn to work directly from $du=(2+2x)dx$. Notice that this is $du=2(x+1)dx$. Do you have the expression $(x+1)dx$ in your integral? – Brian M. Scott Sep 27 '11 at 0:27
@BrianM.Scott you're right, my mistake; thanks. – Davynch0 Sep 27 '11 at 0:30

The question has been fully answered, but I would like to put in an ad for a slightly different substitution, motivated by a desire to avoid square roots.

Let $u^2=2x+x^2$. Then $2u \,du =(2x+x^2)\,dx$, so $(x+1)\,dx=u\,du$. Carry out the substitution. We get $$\int (x+1)\sqrt{2x+x^2}\,dx=\int u^2\,du=\frac{u^3}{3}+C=\frac{(2x+x^2)^{3/2}}{3}+C.$$

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Take $u = x+1,$ then $u^2 - 1 = x^2 + 2 x.$

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but then, I would need to replace $u^2-1$ into $\sqrt {2x+x^2}\;dx$, having now $\sqrt {u^2-1} which is har to integrate; right? – Davynch0 Sep 27 '11 at 0:44 It leads to$\int u\sqrt{u^2-1}\,du$, which is not hard, let$v=u^2-1$. But direct let$u=2x+x^2$does the same thing in one step. – André Nicolas Sep 27 '11 at 1:08 @AndréNicolas So, if I would want to solve$\int (x+1)\sqrt{2x+x^2}$, by substituting$u=x+1$, I would need to substitute twice. Right? – Davynch0 Sep 27 '11 at 1:28 Sort of. After a while you don't need to substitute at all. For example if I really wanted the integral we are talking about, and was not being graded on process, I would "guess" the answer should look like$(2x+x^2)^{3/2}$, because the derivative of$2x+x^2$is basically sitting in front. Then I would mentally differentiate my "guess" and notice the derivative is too big by a factor of$3$. Easy to fix, divide "guess" by$3$. – André Nicolas Sep 27 '11 at 1:44 But if we are going to do things very formally, using the standard integration methods, then yes, if we put$u=x+1$we end up with two substitutions, and if we put$u=2x+x^2$, or$u^2=2x+x^2\$ we end up with one. – André Nicolas Sep 27 '11 at 1:47

$$\frac{du}{dx} = 2 + 2x = 2(x+1),\text{ so } (x+1)\;dx = \frac{du}{2}.$$

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