What is the integral of

dx/sqrt(9+x^2)

and

integral of x^2/(1-3x)dx

To find the integral of dx/sqrt(9+x^2), you can use a trigonometric substitution. Here's how you can do it step by step:

1. Let's start by making a substitution: x = 3 * tan(theta). This substitution helps us get rid of the square root.

2. Solve for dx in terms of d(theta): dx = 3 * sec^2(theta) * d(theta).

3. Substitute dx and x into the original integral: integral of (3 * sec^2(theta) * d(theta)) / sqrt(9 + (3 * tan(theta))^2).

4. Simplify the expression inside the square root: 9 + 9 * tan^2(theta) = 9 * sec^2(theta).

5. Substitute the simplified expression into the integral: integral of (3 * sec^2(theta) * d(theta)) / sqrt(9 * sec^2(theta)).

6. Simplify the expression further: integral of (3 * sec^2(theta) * d(theta)) / (3 * sec(theta)).

7. Cancel out the 3's in the numerator and denominator: integral of sec(theta) * d(theta).

8. The integral of sec(theta) is ln|sec(theta) + tan(theta)| + C, where C is the constant of integration.

9. Substitute back for theta using the original substitution: theta = arctan(x/3).

10. Finally, substitute back into the original variable x to get the answer: ln|sec(arctan(x/3)) + tan(arctan(x/3))| + C.

Now let's move on to the second integral:

For the integral of x^2/(1-3x)dx, we can use polynomial long division to simplify the expression:

1. Divide x^2 by (1-3x). The result is x + 3.

2. Rewrite the integral as: integral of (x + 3) dx/(1-3x).

3. Split the integral into two parts: integral of x dx/(1-3x) + integral of 3 dx/(1-3x).

4. Integrate each part separately.

- For the first part, integral of x dx/(1-3x), you can use the substitution method. Let u = 1-3x, so du = -3dx and dx = -du/3. Substitute these into the integral to get:
-3 integral of (x du/3)/u. Simplifying this gives: - integral of x du/u.

- Integrating - integral of x du/u gives: - (1/2)u^2 + C = - (1/2)(1-3x)^2 + C.

- For the second part, integral of 3 dx/(1-3x), you can directly integrate 3/(1-3x) with respect to x. The integral is 3ln|1-3x|.

5. Combine the two solutions to get the final answer: - (1/2)(1-3x)^2 + 3ln|1-3x| + C, where C is the constant of integration.

And that's how you find the integrals of dx/sqrt(9+x^2) and x^2/(1-3x).