Integration by parts is much easier to do in terms of functions and derivatives rather than differentials, yet the above answerer has chosen the more complicated way.
Integrate the original integrand by parts:
∫ ln(1 - x) dx
Let f'(x) = 1
f(x) = x
Let g(x) = ln(1 - x)
g'(x) = 1 / (x - 1)
∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx
∫ ln(1 - x) dx = xln(1 - x) - ∫ x / (x - 1) dx
Integrate the new integrand by polynomial division:
x / (x - 1) = (x - 1 + 1) / (x - 1)
x / (x - 1) = (x - 1) / (x - 1) + 1 / (x - 1)
x / (x - 1) = 1 + 1 / (x - 1)
∫ x / (x - 1) dx = ∫ [1 + 1 / (x - 1)] dx
∫ x / (x - 1) dx = x + ln|x - 1|
Put it all together to integrate the original function:
dx potential you're integrating with comprehend to x. (in spite of the shown fact that i'm helpful you knew that already). we could locate the antiderivative of x. what happens once you're taking the spinoff of x^2/2? you wind up getting 2x/2 = x. as a result x^2/2 + c is the antiderivative of x. the plus C purely potential that if we've been to tell apart x^2/2 +5 we could nonetheless get purely x. same ingredient if we differentiated x^2/2 +ten thousand we could nonetheless get purely x. so we purely say +C. the guidelines from integrals got here from the guidelines of derivatives. and it as to do with expenditures. (you may look at a calculus e book for the smart purposes. an surprising sort of it got here from physics. i'm hoping this enables it is my first time truly answering a query at here.
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Integration by parts is much easier to do in terms of functions and derivatives rather than differentials, yet the above answerer has chosen the more complicated way.
Integrate the original integrand by parts:
∫ ln(1 - x) dx
Let f'(x) = 1
f(x) = x
Let g(x) = ln(1 - x)
g'(x) = 1 / (x - 1)
∫ f'(x)g(x) dx = f(x)g(x) - ∫ f(x)g'(x) dx
∫ ln(1 - x) dx = xln(1 - x) - ∫ x / (x - 1) dx
Integrate the new integrand by polynomial division:
x / (x - 1) = (x - 1 + 1) / (x - 1)
x / (x - 1) = (x - 1) / (x - 1) + 1 / (x - 1)
x / (x - 1) = 1 + 1 / (x - 1)
∫ x / (x - 1) dx = ∫ [1 + 1 / (x - 1)] dx
∫ x / (x - 1) dx = x + ln|x - 1|
Put it all together to integrate the original function:
∫ ln(1 - x) dx = xln(1 - x) - ∫ x / (x - 1) dx
∫ ln(1 - x) dx = xln(1 - x) - (x + ln|x - 1|)
∫ ln(1 - x) dx = xln(1 - x) - x - ln|x - 1| + C
∫ ln(1 - x) dx = xln(1 - x) - x - ln(1 - x) + C
∫ ln(1 - x) dx = (x - 1)ln(1 - x) - x + C
dx potential you're integrating with comprehend to x. (in spite of the shown fact that i'm helpful you knew that already). we could locate the antiderivative of x. what happens once you're taking the spinoff of x^2/2? you wind up getting 2x/2 = x. as a result x^2/2 + c is the antiderivative of x. the plus C purely potential that if we've been to tell apart x^2/2 +5 we could nonetheless get purely x. same ingredient if we differentiated x^2/2 +ten thousand we could nonetheless get purely x. so we purely say +C. the guidelines from integrals got here from the guidelines of derivatives. and it as to do with expenditures. (you may look at a calculus e book for the smart purposes. an surprising sort of it got here from physics. i'm hoping this enables it is my first time truly answering a query at here.
∫ln(1 - x) dx
u = ln(1 - x)
du = 1/(x - 1) dx
dv = dx
v = x
uv - ∫v du
x*ln(1 - x) - ∫x/(x - 1) dx
x*ln(1 - x) - ∫(1 + 1/(x - 1)) dx
x*ln(1 - x) - x - ln|x - 1| + C