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?

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

3 Answers 3


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.

  • $\begingroup$ 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. $\endgroup$
    – DarkLeader
    Commented Mar 6, 2020 at 1:55
  • $\begingroup$ You have a $t$ at the bottom, right? Distribute it, and take the integral. $\endgroup$
    – Matha Mota
    Commented Mar 6, 2020 at 1:56

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

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


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