Evaluate the integral $\int_0^{1/4}\frac{x-1}{\sqrt{x}-1}\mathrm dx$ so I have this Integral I have to solve without a calculator. 
$$\int_0^{1/4}\dfrac{x-1}{\sqrt{x}-1}\mathrm dx.$$
How would I go about finding the antiderivative of that fraction?
 A: \begin{align}
\int_0^{1/4} \frac{x-1}{\sqrt x-1} \, dx &= \int_0^{1/4} \frac{(\sqrt x-1)(\sqrt x +1)}{\sqrt x-1} \, dx\\
&= \int_0^{1/4} \sqrt x+1 \, dx \\
&= \int_0^{1/4}x^{1/2}+1 \, dx \\
&= \frac23x^{3/2}+x \bigg\vert_0^{1/4} \\
&= \left[\frac23 \left( \frac 14 \right)^{3/2}+\frac 14\right] - 0 \\
&= \frac 13
\end{align}
A: If you follow ganeshie8's hint of $x-1=(\sqrt x-1)(\sqrt x +1)$, then your integral is easier to evaluate:
\begin{align}
\int_0^{1/4} \frac{x-1}{\sqrt x-1} \, dx &= \int_0^{1/4} \frac{(\sqrt x-1)(\sqrt x +1)}{\sqrt x-1} \, dx\\
&= \int_0^{1/4} \sqrt x+1 \, dx \\
&= \int_0^{1/4}x^{1/2}+1 \, dx \\
&= \frac 23x^{3/2}+x \bigg\vert_0^{1/4} \\
&= \left[\frac23 \left( \frac 14 \right)^{3/2}+\frac 14\right] - 0 \\
&= \frac 13
\end{align}
A: Problem:
$\int_0^\frac{1}{4}\frac{x-1}{\sqrt{x}-1}dx$

For the integrand $\frac{x-1}{\sqrt{x}-1}$, substitute $u=\sqrt{x}$ and $du=\frac{1}{2\sqrt{x}}dx$. This gives a new lower bound $u=\sqrt{\frac{1}{4}}=\frac{1}{2}$:
$=2\int_0^\frac{1}{2}\frac{u\left(u^2-1\right)}{u-1}du$

For the integrand $\frac{u\left(u^2-1\right)}{u-1}$, cancel common terms in the numerator and denominator:
$=2\int_0^\frac{1}{2}u\left(u+1\right)du$

For the integrand $u\left(u+1\right)$, substitute $s=u+1$ and $ds=du$. This gives a new lower bound $s=1+0=1$ and upper bound $s=1+\frac{1}{2}=\frac{3}{2}$:
$=2\int_1^\frac{3}{2}s\left(s-1\right)ds$

Expanding the integrand $s\left(s-1\right)$ gives $s^2-s$:
$=2\int_1^\frac{3}{2}\left(s^2-s\right)ds$

Integrate the sum term by term and factor out constants:
$=2\int_1^\frac{3}{2}s^2ds-2\int_1^\frac{3}{2}sds$

Apply the fundamental theorem of calculus. The antiderivative of $s^2$ is $\frac{s^3}{3}$:
$=\frac{2s^3}{3}\Big|_1^\frac{3}{2}-2\int_1^\frac{3}{2}sds$

Evaluate the antiderivative at the limits and subtract. $\frac{2s^3}{3}\Big|_1^\frac{3}{2}=\frac{2}{3}\left(\frac{3}{2}\right)^3-\frac{2*1^3}{3}=\frac{19}{12}$:
$=\frac{19}{12}+\left(-s^2\right)\Big|_1^\frac{3}{2}$

Evaluate the antiderivative at the limits and subtract. $\left(-s^2\right)\Big|_1^\frac{3}{2}=\left(-\left(\frac{3}{2}\right)^2\right)-\left(-1^2\right)=-\frac{5}{4}$:
$=\frac{1}{3}$
