Integrals

**driver327** · August 8th 2007, 09:39 PM

I know the basics of integrals but I am having problems with division with these two problems:
1. indefinite... (x^2+x-2)/(x^2) The answer is x+ln |x| + 2x^-1+C
=x^2+x-2+x^-2
=((x^3/3)+(x^2/2)-(2x)+(x^-1/-1)
=((1/3)(x^3)+(1/2)(x^2)-(2x)-(x^-1) at least my solution. I know I mess this one up.
2. indefinite...((1-x^2)^1/2)-1/(1-x^2)^1/2) The answer is x-(sin(x))^-1+C. Also, I am aware that 1/(1-x^2)^1/2 is (sin x)^-1

**CaptainBlack** · August 8th 2007, 10:38 PM

[quote=driver327;64097]I know the basics of integrals but I am having problems with division with these two problems:
1. indefinite... (x^2+x-2)/(x^2) The answer is x+ln |x| + 2x^-1+C
=x^2+x-2+x^-2
=((x^3/3)+(x^2/2)-(2x)+(x^-1/-1)
=((1/3)(x^3)+(1/2)(x^2)-(2x)-(x^-1) at least my solution. I know I mess this one up.

$\int \frac{x^2+x-2}{x^2}~dx=\int 1+1/x-2/x^2~dx = x+\ln(x)+2/x + C$

RonL

**CaptainBlack** · August 8th 2007, 10:43 PM

Originally Posted by driver327

I2. indefinite...((1-x^2)^1/2)-1/(1-x^2)^1/2) The answer is x-(sin(x))^-1+C. Also, I am aware that 1/(1-x^2)^1/2 is (sin x)^-1

This is ambiguous, but I will assume you mean:

$\int \frac{(1-x^2)^{1/2}-1}{(1+x^2)^{1/2}}~dx=\int 1-\frac{1}{(1+x^2)^{1/2}}~dx$

and the rest you should be able to do.

RonL

**driver327** · August 10th 2007, 06:00 AM

Originally Posted by CaptainBlack

This is ambiguous, but I will assume you mean:

$\int \frac{(1-x^2)^{1/2}-1}{(1+x^2)^{1/2}}~dx=\int 1-\frac{1}{(1+x^2)^{1/2}}~dx$

and the rest you should be able to do.

RonL

On the second part, I see now how it becomes x-sin(x)^-1, but I am a bit confused on how it goes from point A to point B(in the quotes). I know this may be stupid, but i really do not know.

**topsquark** · August 10th 2007, 06:15 AM

Originally Posted by driver327

On the second part, I see now how it becomes x-sin(x)^-1, but I am a bit confused on how it goes from point A to point B(in the quotes). I know this may be stupid, but i really do not know.

It's division. Think of it this way:

Let $b = (1 - x^2)^{1/2}$ . Then you have:

$\frac{b - 1}{b} = \frac{b}{b} - \frac{1}{b} = 1 - \frac{1}{b} = 1 - \frac{1}{(1 - x^2)^{1/2}}$

-Dan

**driver327** · August 10th 2007, 07:05 AM

I must have overlooked that, thanx

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