# Anyway to solve this integral?

1. Aug 13, 2004

### cristina

Int (x+ arcsin (x))/(sqrt(1-x^2))

It's not working when I put U = (x+ arcsin (x)) and dU= 1/(sqrt(1-x^2)
Is there anyway to solve it?

Thanks

2. Aug 13, 2004

### Muzza

If u = x + arcsin(x), then du = (1 + 1/sqrt(1 - x^2))dx. Maybe that's why it "isn't working".

3. Aug 13, 2004

### Galileo

Split the integral and remember (or look up) the derivative of arcsin(x)

4. Aug 13, 2004

### Zurtex

As Galileo says:

$$\int \frac{x + \arcsin x}{\sqrt{1 + x^2}} dx = \int \frac{x}{\sqrt{1 + x^2}} dx + \int \frac{\arcsin x}{\sqrt{1 + x^2}}dx$$

Which makes the former integral of that quite easy, however the latter integral is of the form $f(x) / f'(x)$ which has no standard result. Although it does have a solution it is very long and extremely complex, are you sure this is the integral you were given?

5. Aug 13, 2004

### Galileo

The integrand is $$\frac{\arcsin x}{\sqrt{1 - x^2}}$$
with a minus sign, so the integrand has the form $f(x)f'(x)$

6. Aug 13, 2004

### Zurtex

I'm really sorry, I don't know where my mind went, your totally right.

7. Aug 13, 2004

### cristina

Can I do it with UV - Integra (VdU)

8. Aug 13, 2004

### Zurtex

Your probably can but both the integrals are simple substiutions.

9. Aug 13, 2004

### cristina

can you explain more pls?

10. Aug 13, 2004

### Zurtex

Whenever you have an integral of the form:

$$\int gf(x) * a * f'(x) dx$$

Where 'a' is some constant, then if you use the substitution $u = f(x)$ you get:

$$du = f'(x)dx$$

$$a \int g(u) du$$

You can do this quite easily with both your integrals to get fairly easy integrals.