Find the original function of 2x 1 e x

Solution: Let the original function of the function (-2x+1)e (-x) be y.

y=∫(-2x+1)e^(-x)dx

2x-1)e^(-x)d(-x)

2x-1)d[e (-x)] is obtained using the partial integral formula:

2x-1)e^(-x)-∫e^(-x)d(2x-1)(2x-1)e^(-x)+2∫e^(-x)d(-x)2xe^(-x)-e^(-x)+2e^(-x)+c(2x+1)e^(-x)+c

Therefore, the original function of the function (-2x+1)e (-x) is: (2x+1)e (-x)+c

You are equivalent to parsley to find a normal distribution.

If the function exists, the book will not need to appear its distribution results in the form of a list.

Here's the formula for your reference:

e^(-x^2) =n,0,∞]x^(2 n) (1)^n)/n!

So he dressed up e (-x 2)dx= n,0, ]x (2 n) (1) n) n!dx

n,0,∞]1)^n x^(1 + 2 n))/1 + 2 n) n!)

where [n,0, ]f(n) means that the sum of f(n) is taken from 0 to +, and of course n is only an integer.

e 2xOriginal function: 1 2e 2x+ is a constant.

The analysis process is as follows:

Finding the original function of e 2x is finding the indefinite integral of e 2x.

e^2xdx

1/2∫e^2xd2x

1 2e 2x+c (c is constant).

Formula for indefinite integrals.

1. A dx = ax + c, a and c are constants.

2. Front pin x a dx = x (a + 1)] a + 1) +c, where a is a constant and a ≠ 1

3、∫ 1/x dx = ln|x| +c

4. A x dx = 1 LNA) A x + C, where A > 0 and A ≠ 1

5、∫ e^x dx = e^x + c

6. cosx dx = sinx + c

7. Pure chain sinx dx = cosx + c

8、∫ cotx dx = ln|sinx| +c = ln|cscx| +c

(-2x+1)e^(-x)dx

Huai Sui Cha (-2x+1)d[-e (-x)](2x+1)*[e (-x)]-e (-x)d(-2x+1)2xe (-x)-e (-x)- e (-x)*(2)dx2xe (-x)-e (-x)+2 e (-x)d(-x)2xe (-x)-e (-x)+e (-x)2xe (-x)

The proto-finch function of 2x+1)e (-x) is 2xe (-x)+c

e^(2x)/(1+e^(2x)) dx

1/2)∫ dln(1+e^(2x))

1/2)ln(1+e^(2x)) c

E (2x) (1+e (2x)) = (1 2)ln(1+e (2x)) c