$\int^2_0\dfrac{(2x^{1/2}-1)}{((x+1-2x^{1/2})^{0.5}+x^{0.5})}\,dx$
I felt like I solve this question correctly, but when I checked the answer with my calculator TI-nspire CAS it gave me 4/3 , and wolframapha also gave me 4/3. Isnt it supposed to be 2. Can anyone help me?
The moral of the story here is that $\sqrt{\rm thing^2} = |\rm thing|$, which necessitates some care when you're working where the $\rm thing$ is negative.
Using $u$-substitution with $u = \sqrt{x}$, you do indeed get the integral
$$\int_0^\sqrt2 \frac{2u(2u - 1)\ du}{\sqrt{u^2 - 2u + 1} + u}$$
However, the issue is that $$\sqrt{u^2 - 2u + 1} = \sqrt{(u - 1)^2} = |u - 1| = \begin{cases}u - 1, & u \ge 1 \\ -(u - 1), &u < 1\end{cases}$$
with $u = 1$ in the interval $[0, \sqrt{2}]$ over which you're integrating.
Thus, you'll have to split the integral up into
$$\int_0^\sqrt2 \frac{2u(2u - 1)\ du}{\sqrt{u^2 - 2u + 1} + u} = \int_0^1 \frac{2u(2u - 1)\ du}{-(u - 1) + u} + \int_1^\sqrt{2} \frac{2u(2u - 1)\ du}{u - 1 + u}$$
which does indeed give the expected value of $\frac{4}{3}$.