1. ## Integrals

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

2. [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.

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

RonL

3. 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:

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

and the rest you should be able to do.

RonL

4. Originally Posted by CaptainBlack
This is ambiguous, but I will assume you mean:

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

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.

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

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

-Dan

6. I must have overlooked that, thanx