Indefinite integral of $\frac{\sqrt{x}}{\sqrt{x}+1}$

Question

For this I tried using the substitution technique, but it got me nowhere near the right answer.

What my notepad looks like: $$f(x) = \dfrac{\sqrt{x}}{\sqrt{x}+1}$$ and $$F(x) = \int f(x) = \begin{cases} \sqrt{x} &= t \\ dx &= 2t\cdot dt \end{cases} = \int\frac{t}{t+1}\cdot2t\cdot dt = \boxed{\dfrac{4t(t+1)-2t^2}{(t+1)^2}+c}$$

After swapping $t$ with $x$ again the answer is not even close; where did I go wrong?

Also as a follow up question:

• How do I know when the proper situation is to use integration by substitution above integration by parts?

Answer 1

In answer to the first question, the basic expression inside the OP's boxed formula, $4t(t+1)-2t^2\over(t+1)^2$, is the derivative of $2t^2\over t+1$, not its integral.

As for the second question, there's no hard and fast rule (at least none that I know of), but in this case using a substitution to get rid of the square root signs was a good idea.

Answer 2

Hint. Try instead substituting $t=\sqrt{x}+1$. Then $x=(t-1)^2$, and $\mathrm{d}x=2(t-1)\,\mathrm{d}t$, and hence:

$$\begin{align} \int\frac{\sqrt{x}}{\sqrt{x}+1}\,\mathrm{d}x &=\int\frac{t-1}{t}\cdot2(t-1)\,\mathrm{d}t\\ &=2\int\frac{(t-1)^2}{t}\,\mathrm{d}t\\ \end{align}$$

Answer 3

$$\sqrt x + 1 = t \implies \sqrt x = t-1$$ $$dx =2(t-1)\,dt, \quad \sqrt x = t-1$$

$$\int \frac{2(t-1)^2}t\,dt $$

Answer 4

First:

$$\left(\frac{2t^2+4t}{(t+1)^2}\right)'=\frac{4(t+1)^2-4t(t+2)}{(t+1)^3}=\frac4{(t+1)^3}$$

and this doesn't look like the function you were trying to integrate: $\;\frac{2t^2}{t+1}\;$ !

Your function in the integral is :

$$\frac{2t^2}{t+1}=2t-\frac{2t}{t+1}=2t-2+\frac2{t+1}$$

and thus the integral in fact is

$$\int\left(2t-2+\frac2{t+1}\right)dt=t^2-2t+2\log|t+1|+C$$

and taking back the original variable according to your substitution:

$$t=\sqrt x\implies \int\frac{\sqrt x}{1+\sqrt x}dx=x-2\sqrt x+2\log(\sqrt x+1)+C$$