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The formula for finding primitives by parts is $$\int f(x)g'(x) \ dx= f(x)g(x)-\int f'(x)g(x) \ dx$$ but in reality what we use is the following formula: $$ \int f(x)g(x) \ dx= f(x)G(x) -\int f'(x)G(x) \ dx $$ where $G(x)=\int g(x) \ dx$ (with $c=0$). Why can we do that? Why isn't the constant different from zero?

Is it because if we were to allow a constant different from zero, we would get the same? $$\int f(x)g(x) \ dx=\\ f(x)(G(x)+c_1) -\int f'(x)(G(x)+c_1) \ dx= \\ f(x)G(x) -\int f'(x)G(x) \ dx + c_1(f(x)-\int f'(x) \ dx) = \\ f(x)G(x) -\int f'(x)G(x) \ dx+ c_2=\\ f(x)G(x) -\int f'(x)G(x) \ dx$$

JMP
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An old man in the sea.
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2 Answers2

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The constant can be absorbed by the fact the the 'integration by parts' method takes an indefinite integral and replaces it with a function and another indefinite integral, so the constant term has yet to be realised.

JMP
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1

The method of integration by parts is nothing but the derivative of the product of two functions, reversed.

Indeed you start with

$$\frac{\text{d}}{\text{d}x} f(x)g(x) = f'(x)g(x) + f(x)g'(x)$$

Integrating each term:

$$\int \frac{\text{d}}{\text{d}x} f(x)g(x)\ \text{d}x = \int f'(x)g(x)\ \text{d}x + \int f(x)g'(x)\ \text{d}x$$

And you get the rule of integration by parts.

The constants of integration can be set to zero, or can be set such their sum will be zero. Constants don't matter, because when you derive they will disappear.

$$\frac{\text{d}}{\text{d}x} f(x)g(x) + C = f'(x)g(x) + f(x)g'(x)$$

Et cetera.

Enrico M.
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