# integrate $\int \frac{x}{1+ \sqrt{x}}\ dx$

I'm trying to perform the following integration:

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

so I marked $$t = 1 + \sqrt{x}$$ and $$dt = x + (2/3)x^{3/2} + C$$

I got stuck here after trying to simplify it more or try to break it into two different integrals.

How can I move forward here?

• $dt$ should be the derivative, not the integral, of $t$. So: $dt = (1/2) x^{-1/2} dx$. Commented Mar 6, 2020 at 1:37
• @GEdgar thanks, still stuck though Commented Mar 6, 2020 at 1:42

We can play around with the substitution choice to ensure that the integral is expressed purely in terms of $$t$$.

From the substitution choice we infer: $$t=\sqrt{x}+1$$ $$\implies t-1=\sqrt{x}$$ $$\implies (t-1)^2=x.$$

As GEdgar said, after using the substitution $$t=\sqrt{x}+1$$, you should obtain: $$dt=\frac{1}{2\sqrt{x}}dx$$ $$\implies 2\sqrt{x}dt=2(t-1)=dx.$$

We now make use of these equalities back in the integral:

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

We can expand the numerator and then divide each term by $$t$$, like so:

$$2\int\frac{t^3-3t^2+3t-1}{t}dt$$ $$=2\int t^2-3t+3-\frac{1}{t}dt$$

Can you proceed from here?

You said $$t = 1 + \sqrt{x}$$ gives $$dt = x + 2x^{3/2}/3 + C,$$ which is not correct. It seems instead of taking derivative, you took integral. I am giving you some hinits.

Suppose $$t = 1 + \sqrt{x}$$. Then $$x = (t-1)^2.$$ Hence, $$dx = 2(t-1)dt.$$

Therefore, the integral becomes $$\int \dfrac{(t-1)^2 . 2(t-1)}{t} dt.$$

Note that $$(t-1)^3 = t^3 -3t^2 +3t -1.$$ Hence, we essentillay have the following thing:

$$\int 2(\dfrac {t^3 -3t^2 +3t -1}{t}) dt$$

= $$2\int( t^2 -3t +3 - \frac{1}{t})dt.$$

Do the rest of the steps by yourself.

• ok I got to $$\int (2(t-1)^3)/t$$ I still fail to understand how to perform it, note that I just learned integrals, it might be really simple but I don't know any techniques for it. Commented Mar 6, 2020 at 1:55
• You have a $t$ at the bottom, right? Distribute it, and take the integral. Commented Mar 6, 2020 at 1:56

You can also proceed with the substitution $$x = t^2, dx = 2t \ \mathrm{d} t$$:

$$\int \dfrac{t^2}{1+t} \ 2t \ \mathrm{d} t = 2 \int \dfrac{t^3}{1+t} \mathrm{d} t = 2 \int \dfrac{t^2(t+1)-t(t+1)+(t+1)-1}{1+t} \ \mathrm{d} t$$ $$= 2 \left(\frac{t^3}{3} -\frac{t^2}{2}+t-\ln|1+t|+ C\right)$$ $$= \frac{2x \sqrt{x}}{3} -x+2\sqrt{x}-2\ln|1+\sqrt{x}|+ C_1$$