Missing coefficient in $u$-substitution

Question

I need to integrate the following: $$\int \dfrac{\sqrt x}{1+\sqrt x}\, dx$$
So I do $u$-substitution with $u = 1+\sqrt{x}$ and $du=\dfrac{1}{\sqrt{x}}\,dx$ and I multiply the numerator and the denominator by $\sqrt{x}$ to get:
$$\int \dfrac{(u-1)^2}{u}du.$$

But according to this website I still need to multiply the result by $2$. Why is it?

Answer 1

In order to solve the integral $$\int \frac{\sqrt x}{1+\sqrt x} dx$$ by $u$ substitution, first you should pick a suitable $u$, as you say, it should be $u=1+\sqrt{x}$. Now the second step is complete the differential in the integral, a basic calculation shows that $du=\frac{1}{2\sqrt{x}}dx$ . Finally, to complete the differential, we should multiply by $1$ our integral, in this case, $1=\frac{1}{2\sqrt{x}}\cdot 2 \sqrt{x}$ for a non zero $x$.

Now rewriting our integral.

$$\int \frac{\sqrt x}{1+\sqrt x} dx=\int \frac{\sqrt x}{1+\sqrt x} dx \cdot \frac{1}{2\sqrt{x}}\cdot 2 \sqrt{x}=2\int \frac{x}{1+\sqrt{x}} \cdot \frac{1}{2\sqrt{x}}dx$$

Finally, since $u=1+\sqrt{x}$, we get that $x=(u-1)^2$. So by a simple substitution, we get

$$\int \frac{\sqrt x}{1+\sqrt x} dx=2\int \frac{(u-1)^2}{u} du $$.

Answer 2

$$\int \frac{\sqrt x}{1+\sqrt x} dx= \ldots$$ With the $u$-substitution with $u = 1+\sqrt{x}$ you have

$$du=\frac{dx}{2\sqrt x} \iff dx=2\sqrt x\, du$$

Hence

$$\int \frac{\sqrt x}{1+\sqrt x} dx=\color{red}{2}\int \frac{(u-1)^2}{u}du$$

Expand the integranda function we have:

$$\int \frac{(u-1)^2}{u}du=\int\left(u+\frac 1u-2\right)du$$

Apply linearity $$\int u du + \int \frac 1u du -2\int du=\frac{u^2}{2}+\ln u-2u +k, \quad k\in \mathbb R$$

After multiplying by two

$$u^2+2\ln u-4u +k' \tag 1$$

Undo substitution $u = 1+\sqrt{x}$ in $(1)$ you have the thesis.

Answer 3

$$ \displaystyle \begin{array}{rl} \text { Let } u&=\sqrt{x}, \text { then } 2 u d u=d x \\ I = & \displaystyle \int \dfrac{u}{1+u} \cdot 2 u d u \\ = & 2 \displaystyle \int \dfrac{u^2-1+1}{1+u} d u \\ = & 2 \displaystyle \int\left(u-1+\dfrac{1}{1+u}\right) d u \\ = & (u-1)^2+2 \ln (1+u)+c \\ = & x-2 \sqrt{x}+2 \ln (1+\sqrt{x})+c \end{array} $$