1.) ∫ (2)/(x-4) dx

2.) ∫ sec^2x tanx dx

3.) ∫ 2 csc^2 xdx

4.) ∫ (3) / sqtr(x+3) dx

5.) ∫ (2x-1) / (x^2 - x)

5.) �ç (2x-1) dx / (x^2 - x)

let x^2 -x = u
du = 2x -1

Integral = �ç du/u = ln u = ln(x^2-u)

1.) To find ∫ (2)/(x-4) dx, we can use the method of u-substitution.

Let u = x - 4. Then, du = dx. Rearranging, we have dx = du.

So, the integral becomes ∫ (2)/(u) du.

Integrating ∫ (2)/(u) du gives us 2 ln |u| + C, where C is the constant of integration.

Substituting back in x, we have 2 ln |x - 4| + C as the final answer.

2.) To find ∫ sec^2x tanx dx, we can use the method of u-substitution.

Let u = secx. Then, du = secx tanx dx. Rearranging, we have dx = du / (secx tanx).

So, the integral becomes ∫ du.

Integrating ∫ du gives us u + C, where C is the constant of integration.

Substituting back in u = secx, we have secx + C as the final answer.

3.) To find ∫ 2 csc^2xdx, we can use the method of u-substitution.

Let u = sinx. Then, du = cosx dx. Rearranging, we have dx = du / cosx.

So, the integral becomes ∫ (2) / (u^2) du.

Integrating ∫ (2) / (u^2) du gives us -2/u + C, where C is the constant of integration.

Substituting back in u = sinx, we have -2/sinx + C as the final answer.

4.) To find ∫ (3) / sqrt(x+3) dx, we can use the method of u-substitution.

Let u = x + 3. Then, du = dx. Rearranging, we have dx = du.

So, the integral becomes ∫ (3) / sqrt(u) du.

Integrating ∫ (3) / sqrt(u) du gives us 6 sqrt(u) + C, where C is the constant of integration.

Substituting back in u = x + 3, we have 6 sqrt(x + 3) + C as the final answer.

5.) To find ∫ (2x-1) / (x^2 - x) dx, we can use the method of partial fractions.

First, we factorize the denominator: x^2 - x = x(x - 1).

Next, we express the integrand as a sum of two fractions: A/x + B/(x - 1).

To find A and B, we clear the fractions by multiplying both sides of the equation by (x)(x - 1):

2x - 1 = A(x - 1) + B(x).

Expanding and rearranging the equation, we get:

2x - 1 = Ax - A + Bx.

Now, equating the coefficients of the corresponding powers of x, we have:

2 = A + B (for the coefficient of x)
-1 = -A (for the constant term)

Solving these equations, we find:
A = 1
B = 1

So, the integral becomes ∫ (1/x) + (1/(x - 1)) dx.

Integrating each term, we get ln|x| + ln|x - 1| + C, where C is the constant of integration.

Combining the logarithms using properties of logarithms, we have ln|x(x - 1)| + C as the final answer.