Integration of Irrational Functions

Integration of Irrational Functions
Suppose that is a rational function of and . Then
1. $\displaystyle{\int R(x,\sqrt[n]{\rm linear\ polynomial })dx}$
Let $t = \sqrt{\rm linear\ polynomial}$ . Then we can get integration of rational function.
[2] $\int R(x,\sqrt{\rm quadratic\ polynomial})dx, \ {\rm where}\ a \neq 0$
1. after completing the square. Then let $x = a\tan{t}$

$\displaystyle dx = a \sec^{2}{t}dt, x^2 + a^2 = a^2(\tan^{2}{t} + 1) = a^2 \sec^{2}{t}$

2. after completing the square. Then let $x = a\sin{t}$

$\displaystyle dx = a\cos{t}dt, a^2 -x^2 = a^2(1 - \cos^{2}{t}) = a^2\sin^{2}{t}$

3. after completing the square, Then let $x = a\sec{t}$

$\displaystyle dx = a\sec{t}\tan{t}dt, x^2 - a^2 = a^2(\sec^{2}{t} - 1) = a^2 \tan^{2}{t}$

Example 3..12 Find the following integral.
1. $\displaystyle{\int \frac{dx}{1 - \sqrt{x}}}$ 2. $\displaystyle{\int \frac{xdx}{\sqrt{x^2 - 4}}}$
SOLUTION 1. Let $\sqrt{x} = t$ . Then and . Thus

$\displaystyle \int \frac{dx}{1 - \sqrt{x}}$ $\displaystyle =$ $\displaystyle \int \frac{2t}{1 - t}dt = \int \frac{-2(1-t)+2}{1-t}\:dt = \int (-2 + \frac{2}{1-t})dt$

$\displaystyle =$ $\displaystyle -2t - 2\log{\vert 1 - t\vert} + c = - 2\sqrt{x} - 2\log{\vert 1 - \sqrt{x}\vert} + c \ensuremath{\ \blacksquare}$

2. Inside of the squareroot is the form of . Let $x = 2\sec{t}$ . Then $dx = 2\sec{t}\tan{t}dt$ . Note that $\sqrt{x^2 - 4} = \sqrt{4\sec^2{x} - 4} = \sqrt{4(\sec^{2}{x} - 1)} = 2\tan{t}$ . Thus

$\displaystyle \int \frac{x}{\sqrt{x^2 - 4}}dx$ $\displaystyle =$ $\displaystyle \int \frac{2\sec{t}}{2\tan{t}}\cdot 2\sec{t}\tan{t}dt = 2 \int \sec^{2}{t}dt$

$\displaystyle =$ $\displaystyle 2 \tan{t} + c = \sqrt{x^2 - 4} + c \ensuremath{\ \blacksquare}$

Exercise 3..12 Find the following integrals.
1. $\displaystyle{\int \frac{dx}{x\sqrt{x^2 - 4}}}$ 2. $\displaystyle{\int{\sqrt{6x - x^2 - 8}}\ dx}$
SOLUTION 1. Let $x = 2\sec{t}$ . Then $dx = 2\sec{t}\tan{t}dt$ , $\sqrt{x^2 - 4} = 2\tan{t}$ . Thus,

$\displaystyle \int \frac{dx}{x\sqrt{x^2 -4}}$ $\displaystyle =$ $\displaystyle \int \frac{2\sec{t}\tan{t}dt}{4\sec{t}\tan{t}} = \frac{1}{2}\int dt = \frac{1}{2}t + c$

$\displaystyle =$ $\displaystyle \frac{1}{2}\tan^{-1}\frac{\sqrt{x^2 - 4}}{2} + c\ensuremath{\ \blacksquare}$

2. Completing the square,

$\displaystyle 6x - x^2 - 8 = 9 - 9 + 6x - x^2 - 8 = 1 - (9 - 6x + x^2) = 1 - (3-x)^2.$
Now it is in the form of . Thus conside the right triangle with the hypotenuse , the opposite side of the angle is .
Let $3-x = \sin{t}$ . $-dx = \cos{t}dt$ .

$\displaystyle \sqrt{1 - (3-x)^2} = \sqrt{1 - \sin^{2}{t}} = \sqrt{\cos^{2}{t}} = \cos{t}$

$\displaystyle \int{\sqrt{6x - x^2 - 8}}\ dx$ $\displaystyle =$ $\displaystyle \int{\cos{t}(-\cos{t})}\ dt$

$\displaystyle =$ $\displaystyle -\int{\cos^{2}{t}}\ dt$

$\displaystyle =$ $\displaystyle -\int{\frac{1 + \cos{2t}}{2}}\ dt = -[\frac{t}{2} + \frac{\sin{2t}}{4}] + c$

$\displaystyle =$ $\displaystyle -[\frac{t}{2} + \frac{\sin{t}\cos{t}}{2}] + c$

$\displaystyle =$ $\displaystyle -\frac{\sin^{-1}{(3-x)}}{2} - \frac{(3-x)\sqrt{1 - (3-x)^2}}{2} + c\ensuremath{\ \blacksquare}$

Exercise A

1.

Workout the following integrals?D
(a) $\displaystyle{\int{\frac{1}{1+\sqrt{x}}}\ dx}$ (b) $\displaystyle{\int{\frac{\sqrt{x}}{\sqrt{x} + 1}}\ dx}$ (c) $\displaystyle{\int{\frac{dx}{\sqrt{1 - e^x}}}}$ (d) $\displaystyle{\int{\frac{x}{\sqrt{x - 1}}}\ dx}$ (e) $\displaystyle{\int{\frac{x}{\sqrt{x^2 + 4}}}\ dx}$ (f) $\displaystyle{\int{\frac{x^{3}}{\sqrt{x^2 + 4}}}\ dx}$ (g) $\displaystyle{\int{\frac{dx}{\sqrt{x^2 - 2x - 3}}}}$
(h) $\displaystyle{\int{\frac{\sqrt{x^{2}-1}}{x}}\ dx}$

Exercise B

1.
Work out the following integrals?D

(a) $\displaystyle{\int{x\sqrt{1+x}} \ dx}$ (b) $\displaystyle{\int{\frac{\sqrt{x}}{\sqrt{x} - 1}}\ dx}$ (c) $\displaystyle{\int{\frac{dx}{\sqrt{1 + e^x}}}}$ (d) $\displaystyle{\int{x^2 \sqrt{x - 1}}\ dx}$

(e) $\displaystyle{\int{\sqrt{\frac{x+1}{x-1}}}\ dx}$
2.
Work out the following integrals?D

(a) $\displaystyle{\int{\frac{x}{\sqrt{x^2 - 4}}}\ dx}$ (b) $\displaystyle{\int{\frac{x^2}{\sqrt{4 - x^2}}}\ dx}$ (c) $\displaystyle{\int{\frac{e^x}{9 - e^{2x}}}\ dx}$ (d) $\displaystyle{\int{\frac{\sqrt{1 - x^2}}{x^4}}\ dx}$ (e) $\displaystyle{\int{\frac{dx}{x^2\sqrt{x^2 - a^2}}}}$ (f) $\displaystyle{\int{\frac{dx}{e^x\sqrt{4 + e^{2x}}}}}$ (g) $\displaystyle{\int{\frac{dx}{\sqrt{x^2 - 2x - 3}}}}$

(h) $\displaystyle{\int{\frac{x}{\sqrt{6x - x^2}}}\ dx}$ (i) $\displaystyle{\int{\frac{x}{\sqrt{x^2 - 2x - 3}}}\ dx}$ (j) $\displaystyle{\int{\sqrt{6x - x^2 - 8}}\ dx}$

(k) $\displaystyle{\int{x\sqrt{x^2 + 6x}}\ dx}$

Next:Definite Integrals Up:Integration Previous:Integration of Trigonometric Functions Contents Index

$\displaystyle \int \frac{dx}{1 - \sqrt{x}}$	$\displaystyle =$	$\displaystyle \int \frac{2t}{1 - t}dt = \int \frac{-2(1-t)+2}{1-t}\:dt = \int (-2 + \frac{2}{1-t})dt$
	$\displaystyle =$	$\displaystyle -2t - 2\log{\vert 1 - t\vert} + c = - 2\sqrt{x} - 2\log{\vert 1 - \sqrt{x}\vert} + c \ensuremath{\ \blacksquare}$

$\displaystyle \int \frac{x}{\sqrt{x^2 - 4}}dx$	$\displaystyle =$	$\displaystyle \int \frac{2\sec{t}}{2\tan{t}}\cdot 2\sec{t}\tan{t}dt = 2 \int \sec^{2}{t}dt$
	$\displaystyle =$	$\displaystyle 2 \tan{t} + c = \sqrt{x^2 - 4} + c \ensuremath{\ \blacksquare}$

$\displaystyle \int \frac{dx}{x\sqrt{x^2 -4}}$	$\displaystyle =$	$\displaystyle \int \frac{2\sec{t}\tan{t}dt}{4\sec{t}\tan{t}} = \frac{1}{2}\int dt = \frac{1}{2}t + c$
	$\displaystyle =$	$\displaystyle \frac{1}{2}\tan^{-1}\frac{\sqrt{x^2 - 4}}{2} + c\ensuremath{\ \blacksquare}$

$\displaystyle \int{\sqrt{6x - x^2 - 8}}\ dx$	$\displaystyle =$	$\displaystyle \int{\cos{t}(-\cos{t})}\ dt$
	$\displaystyle =$	$\displaystyle -\int{\cos^{2}{t}}\ dt$
	$\displaystyle =$	$\displaystyle -\int{\frac{1 + \cos{2t}}{2}}\ dt = -[\frac{t}{2} + \frac{\sin{2t}}{4}] + c$
	$\displaystyle =$	$\displaystyle -[\frac{t}{2} + \frac{\sin{t}\cos{t}}{2}] + c$
	$\displaystyle =$	$\displaystyle -\frac{\sin^{-1}{(3-x)}}{2} - \frac{(3-x)\sqrt{1 - (3-x)^2}}{2} + c\ensuremath{\ \blacksquare}$