Find the following indefinite integral: (√x-2ex+9x)dx ○x²-2x+63x²+c 3 3 2 9 -2ex+x+c 7 ○x²-20*+63x²+c 2e 7 3 97 x²-2x+x+c

It seems like there's a bit of confusion in your question. However, I understand that you want to find the indefinite integral of the function (√x - 2e^x + 9x).

The integral of a sum of functions is the sum of the integrals of the functions. So, we can break this problem down into three parts:

1. ∫√x dx
2. ∫-2e^x dx
3. ∫9x dx

Let's solve each part:

1. ∫√x dx = ∫x^(1/2) dx = (2/3)x^(3/2)
2. ∫-2e^x dx = -2e^x
3. ∫9x dx = (9/2)x^2

Adding these together, the indefinite integral of the function (√x - 2e^x + 9x) is:

(2/3)x^(3/2) - 2e^x + (9/2)x^2 + C

where C is the constant of integration.

It seems like there's a bit of confusion in your question. However, I understand that you want to find the indefinite integral of the function (√x - 2e^x + 9x).

The integral of a sum of functions is the sum of the integrals of the functions. So, we can break this problem down into three parts:

1. ∫√x dx
2. ∫-2e^x dx
3. ∫9x dx

Let's solve each part:

1. ∫√x dx = ∫x^(1/2) dx = (2/3)x^(3/2)
2. ∫-2e^x dx = -2e^x
3. ∫9x dx = (9/2)x^2

Adding these together, the indefinite integral of the function (√x - 2e^x + 9x) is:

(2/3)x^(3/2) - 2e^x + (9/2)x^2 + C

where C is the constant of integration.