Can we find $\int 1/(\sqrt{x+2} - \sqrt{x+1}) dx$?

Can we find $\int 1/(\sqrt{x+2} - \sqrt{x+1}) dx$?

Question

1. By using the substitution $x = \cos \theta$, or otherwise, evaluate

$$\begin{equation*} \int_0^1 \frac{1}{(1-x)^{1/2}(1+x)^{3/2}} \:dx. \end{equation*}$$

---

1. Find

$$\begin{align*} \text{(i)}\; &\int \frac{1}{\sqrt{x+2} - \sqrt{x+1}} \:dx, \\ \text{(ii)}\; &\int \frac{x}{(x+1)(x+2)^2} \:dx. \end{align*}$$

Solution

If

$x = \cos \theta$

, then

$dx = -\sin \theta \:d\theta$

, and so

$$\begin{align*} \int_0^1 \frac{1}{(1-x)^{1/2}(1+x)^{3/2}} \:dx &= \int_{\pi/2}^0 \frac{-\sin \theta}{(1 - \cos \theta)^{1/2}(1 + \cos \theta)^{3/2}} \:d\theta \\ &= \int_0^{\pi/2} \frac{\sin \theta}{(1 - \cos \theta)^{1/2}(1 + \cos \theta)^{3/2}} \:d\theta \\ &= \int_0^{\pi/2} \frac{\sin \theta}{(1 - \cos \theta)^{1/2}(1 + \cos \theta)^{1/2}(1 + \cos \theta)} \:d\theta \\ &= \int_0^{\pi/2} \frac{\sin \theta}{(1 - \cos^2 \theta)^{1/2}(1 + \cos \theta)} \:d\theta \\ &= \int_0^{\pi/2} \frac{d\theta}{1 + \cos \theta}. \end{align*}$$

From the

double angle formula for cosine

, we have that

$$\begin{equation*} \cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = 2 \cos^2 \frac{\theta}{2} - 1 \implies 1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}. \end{equation*}$$

Hence,

$$\begin{align*} \int_0^{\pi/2} \frac{d\theta}{1 + \cos \theta} &= \frac{1}{2} \int_0^{\pi/2} \sec^2 \frac{\theta}{2} \:d\theta\\ &=\left[\tan \dfrac{\theta}{2}\right]_0^{\pi/2}\\ &= \tan \frac{\pi}{4} - \tan 0 = 1. \end{align*}$$

1. Find

$$\begin{align*} \text{(i)}\; &\int \frac{1}{\sqrt{x+2} - \sqrt{x+1}} \:dx, \\ \end{align*}$$

Rationalising the denominator, we have that

$$\begin{align*} \int \frac{1}{\sqrt{x+2} - \sqrt{x+1}} \:dx &= \int \frac{\sqrt{x+2} + \sqrt{x+1}}{(\sqrt{x+2} - \sqrt{x+1})(\sqrt{x+2} + \sqrt{x+1})} \:dx \\ &= \int \frac{\sqrt{x+2} + \sqrt{x+1}}{(x+2) - (x+1)} \:dx \\ &= \int \sqrt{x+2} + \sqrt{x+1} \:dx \\ &= \frac{(x+2)^{3/2}}{\frac{3}{2}} + \frac{(x+1)^{3/2}}{\frac{3}{2}} + c \\ &= \frac{2}{3} \left( (x+2)^{3/2} + (x+1)^{3/2} \right) + c \end{align*}$$

where $c$ is a constant of integration.

$$\begin{align*} \text{(ii)}\; &\int \frac{x}{(x+1)(x+2)^2} \:dx. \end{align*}$$

We can attempt to rewrite the integrand using partial fractions: that is, we would hope that there are constants

$A$

,

$B$

, and

$C$

such that

$$\begin{equation*} \frac{x}{(x+1)(x+2)^2} = \frac{A}{x+1} + \frac{B}{x+2} +\frac{C}{(x+2)^2}= \frac{A(x+2)^2 + B(x+1)(x+2)+C(x+1)}{(x+1)(x+2)^2}. \end{equation*}$$

Thus $x = A(x+2)^2 + B(x+1)(x+2)+C(x+1)$ for all values of $x$.

Putting $x = -1$, we have $A = -1$, while putting $x = -2$ gives $C=2$ and putting $x = 0$ yields $B = 1$.

As a result,

$$\begin{align*} \frac{x}{(x+1)(x+2)^2} &= \frac{1}{x+2} + \frac{2}{(x+2)^2} - \frac{1}{x+1} \end{align*}$$

and so

$$\begin{align*} \int \frac{x}{(x+1)(x+2)^2} \:dx &= \int \frac{1}{x+2} + \frac{2}{(x+2)^2} - \frac{1}{x+1} \:dx \\ &= \ln\big\vert x+2\big\vert - \frac{2}{x+2} - \ln\big\vert x+1\big\vert + c \end{align*}$$

where $c$ is a constant of integration.