# 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?

• hint : $x-1 = (\sqrt{x}-1)(\sqrt{x}+1)$ Commented Sep 21, 2014 at 18:40

\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}

• Actually I posted this when yours was incorrect
– KFC
Commented Sep 21, 2014 at 18:54
• Oh, my bad, I didn't know you were the OP posting this. Sorry. Commented Sep 21, 2014 at 18:55
• Yeah, I see that you fixed your post. So, I'll still select yours, no worries.
– KFC
Commented Sep 21, 2014 at 18:57

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}

• Yeah, I got 1/3 for my final answer.
– KFC
Commented Sep 21, 2014 at 18:51
• The evaluation would not be different if the interval included $1$, it really is not an improper integral. Commented Sep 21, 2014 at 18:55

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. $\left(-s^2\right)\Big|_1^\frac{3}{2}=\left(-\left(\frac{3}{2}\right)^2\right)-\left(-1^2\right)=-\frac{5}{4}$:
$=\frac{1}{3}$