de-CH
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math
Partial Integration
i-07-03
expression
3
[ ["x^2", "-\\cos(x)", "2x", "\\sin(x)", "x^2\\sin(x)", "-2x\\cos(x)", "-x^2\\cos(x) + 2x\\sin(x) + 2\\cos(x)", "- (2x\\sin(x) + 2\\cos(x))"], ["x^2", "\\sin(x)", "2x", "\\cos(x)", "x^2\\cos(x)", "2x\\sin(x)", "x^2\\sin(x) + 2x\\cos(x) - 2\\sin(x)", "-2x\\cos(x) + 2\\sin(x)"], ["x^2", "e^x", "2x", "e^x", "x^2e^x", "2xe^x", "x^2e^x - 2xe^x + 2 e^x", "2xe^x - 2 e^x"] ] randRange(0,functionBank.length-1) functionBank[fNum]

Determine \displaystyle \int f[4] \; dx.

Use C as the integration constant.

f[6] + C

For partial integration, we use that

\displaystyle \int f(x)g'(x) \; dx = f(x)g(x) - \int f'(x)g(x)\; dx + C.

An appropriate choice of f and g should ensure to find a primitive of f'(x)g(x) in an easier way than as for f(x)g'(x).

Here, suitable choices are \displaystyle f(x) = f[0] and \displaystyle g'(x) = f[3].

With \displaystyle f'(x) = f[2] and \displaystyle \color{blue}{g(x) = f[1]}, it is

\displaystyle f'(x){\color{blue}g(x)} = f[2] \cdot \left(f[1] \right) = f[5] .

(In \displaystyle \int g'(x) \; dx = g(x) is the integration constant C=0.)

Verify

\displaystyle \int f[5] \; dx = f[7] + C

with another partial integration.

It is \displaystyle f(x){\color{blue}g(x)} = f[0] \cdot \left( {\color{blue}f[1]} \right).

Thus, together

\displaystyle \int f[4] \; dx = f[6] + C.