Calculating the integral $\frac{x}{1 + \sqrt{x}}$

Question

I am trying to calculate the following integral: $$ \int_0^1 \frac{x}{1 + \sqrt{x}} $$ I have tried a subtitution of $u = \sqrt{x}$, gives $x = u^2$ and $dx/du = 2u$, so $dx = 2u du$. But then I get $$ 2 \int_0^1 \frac{u^3}{1 + u} $$ Still no idea how to proceed. I have thought about using $\ln(x)$ or $\arctan(x)$ but I'm stuck. Anyone knows how to calculate it?

Answer 1

Hint:

Start with $$\sqrt x+1=u\implies x=(u-1)^2,dx=2(u-1)\ du$$

Answer 2

\begin{align*} \dfrac{u^{3}}{u+1}&=\dfrac{(u+1-1)^{3}}{u+1}\\ &=\dfrac{(u+1)^{3}-3(u+1)^{2}+3(u+1)-1}{u+1}\\ &=(u+1)^{2}-3(u+1)+3-\dfrac{1}{u+1}, \end{align*} then \begin{align*} \int_{0}^{1}(u+1)^{2}du=\dfrac{1}{3}(u+1)^{3}\bigg|_{u=0}^{u=1}=\cdots \end{align*} and the last one is \begin{align*} \int_{0}^{1}\dfrac{1}{u+1}du=\log(u+1)\bigg|_{u=0}^{u=1}=\cdots \end{align*}

Answer 3

Just make the division of $x^3$ by $1+u$. You'll obtain $$\int_0^1 p(u)\,\mathrm du+\int_0^1 \frac 1{1+u}\,\mathrm du,$$ where $p(u)$ is a quadratic polynomial.