Find the anti derivative of the following:

1. u^-2(1-u^2+u^4)du
2. (sqrt(x)+ 3/x - 2e^(x))dx

The anti derivative symbol is in front of the equations.

Also need help on find f(x) so that it satisfies f'(x) = 1+e^x+1/x and f(1) = 3+e.

Please show steps so I can understand. Thanks.

To find the antiderivative of each equation, we can use the rule of integration. The general method is to apply the power rule, constant multiple rule, and sum/difference rule in reverse order.

1. u^-2(1 - u^2 + u^4)du:
In this case, we can rewrite the equation as follows:
u^-2 - u^0 + u^2du

Now, we can apply the power rule in reverse order:
(u^-2+1) / (-2+1) - (u^0+1) / (0+1) + (u^2+1) / (2+1) + C

Simplifying the equation, we get:
-1/u + 1/u^2 + u^3/3 + C

2. (sqrt(x) + 3/x - 2e^x)dx:
We have three terms in this equation, so we will integrate each term separately.

For the first term, √x, we can use the power rule in reverse order:
(2/3)x^(3/2)

For the second term, 3/x, we can rewrite it as:
3x^-1

To solve it, we use the power rule in reverse:
3 ln|x|

For the third term, -2e^x, the antiderivative is:
-2e^x

Putting all the terms together, we get:
(2/3)x^(3/2) + 3ln|x| - 2e^x + C

Now, let's move on to finding f(x) such that f'(x) = 1 + e^x + 1/x and f(1) = 3 + e.

To find f(x), we will perform the reverse process of differentiation, which is integration. We integrate the expression 1 + e^x + 1/x with respect to x.

∫(1 + e^x + 1/x)dx = ∫dx + ∫e^xdx + ∫(1/x)dx

The first integral is simply x.

The integral of e^x is e^x.

The integral of 1/x can be found by using the natural logarithm:
∫(1/x)dx = ln|x|

Putting it all together, we have:
f(x) = x + e^x + ln|x| + C

To find the value of C, we will use the given condition f(1) = 3 + e:
f(1) = 1 + e + ln|1| + C = 3 + e

Simplifying the equation, we get:
2 + e + C = 3 + e

Therefore, C = 1.

Finally, the function f(x) that satisfies f'(x) = 1 + e^x + 1/x, and f(1) = 3 + e, is:
f(x) = x + e^x + ln|x| + 1