Question

ЛЮДИ ДОБРЫЕ!!!ПОМОГИТЕ!!!ПОЖАЛУЙСТА!!!!ОЧЕНЬ НАДО!!!!Найти первообразную для функции:- (x + 1)

Answer 1

По частям:
$-\int e^{-x}(x+1)\,dx=\int(x+1)\,de^{-x}=(x+1)\,e^{-x}-\int e^{-x}\,d(x+1)=\\=(x+1)\,e^{-x}-\int e^{-x}\,dx=(x+1)\,e^{-x}+e^{-x}+C=(x+2)\,e^{-x}+C$

Answer 2

Finding the Indefinite Integral of (x + 1)

To find the indefinite integral of the function \(x + 1\), we can use the following steps:

1. Add 1 to the exponent of x and divide by the new exponent. 2. Add the constant of integration, denoted by "C".

The indefinite integral of \(x + 1\) is:

\[ \int (x + 1) \, dx = \frac{x^2}{2} + x + C \]

Where \(C\) is the constant of integration.

This result is obtained by applying the power rule of integration, which states that the integral of \(x^n\) with respect to \(x\) is \(\frac{x^{n+1}}{n+1}\) plus the constant of integration.

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Therefore, the indefinite integral of \(x + 1\) is \(\frac{x^2}{2} + x + C\).