Integration of Irrational Functions

Integration of Irrational Functions

Suppose that $R(x,y)$ is a rational function of $x$ and $y$. Then

1. $${\int R(x,\sqrt[n]{\rm linear\ polynomial })dx}$$
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. $x^2 + a^2$ after completing the square. Then let $x = a\tan{t}$

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

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

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

3. $x^2 - a^2$ after completing the square, Then let $x = a\sec{t}$

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

Example 3..12   Find the following integral.
1. $${\int \frac{dx}{1 - \sqrt{x}}}$$ 2. $${\int \frac{xdx}{\sqrt{x^2 - 4}}}$$

SOLUTION 1. Let $\sqrt{x} = t$. Then $x = t^2$ and $dx = 2t dt$. Thus

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

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

2. Inside of the squareroot is the form of $x^2 - a^2$. 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

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

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

Exercise 3..12   Find the following integrals.
1. $${\int \frac{dx}{x\sqrt{x^2 - 4}}}$$ 2. $${\int{\sqrt{6x - x^2 - 8}}\ dx}$$

SOLUTION 1. Let $x = 2\sec{t}$. Then $dx = 2\sec{t}\tan{t}dt$, $\sqrt{x^2 - 4} = 2\tan{t}$. Thus,

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

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

2. Completing the square,

$$6x - x^2 - 8 = 9 - 9 + 6x - x^2 - 8 = 1 - (9 - 6x + x^2) = 1 - (3-x)^2.$$

Now it is in the form of $a^2 - x^2$. Thus conside the right triangle with the hypotenuse $1$, the opposite side of the angle $t$ is $3-x$.

Let $3-x = \sin{t}$. $-dx = \cos{t}dt$.

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

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

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

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

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

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

Exercise A

1.

Workout the following integrals?D

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

(h) $${\int{\frac{\sqrt{x^{2}-1}}{x}}\ dx}$$

Exercise B

1.

Work out the following integrals?D

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

(e) $${\int{\sqrt{\frac{x+1}{x-1}}}\ dx}$$

2.

Work out the following integrals?D

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

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

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