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# Integration by parts Tweet

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1. Integration by parts
I've got to integrate ln(1+x) using integration by parts, but I don't know what f' and g are. Any help?
2. you can take f' as 1 and g as ln(1+x) i think.
3. (Original post by Acaila)
I've got to integrate ln(1+x) using integration by parts, but I don't know what f' and g are. Any help?
Int. ln(1 + x) dx. = Int. 1.ln(1 + x) dx.
Let u = ln(1 + x) -> du/dx = 1/(1 + x)
Let v' = 1 - > v = x

-> Int. ln(1 + x) dx. = x.ln(1 + x) - Int. x/(1 + x)

x/(1 + x) = A + B/(1 + x) = [A(1 + x) + B]/(1 + x)

Equating Numerators:

-> A(1 + x) + B = x
Let x = -1: B = -1
Let x = 0: A - 1 = 0 -> A = 1

Hence: x/(1 + x) = 1 - 1/(1 + x)

Hence: Int. ln(1 + x) dx. = x.ln(1 + x) - Int. 1 - 1/(1 + x) = x.ln(1 + x) - [x - ln|1 + x| + c] = x.ln(1 + x) - x + ln(1 + x) + k

-> Int. ln(1 + x) dx. = ln(1 + x)[x + 1] - x + k
4. (Original post by Acaila)
I've got to integrate ln(1+x) using integration by parts, but I don't know what f' and g are. Any help?
f' = 1, g = ln(1 + x). You want this because you can differentiate ln(1 + x) easily, as well as being able to integrate 1.
5. (Original post by Acaila)
I've got to integrate ln(1+x) using integration by parts, but I don't know what f' and g are. Any help?
Int f'g = fg - Int fg'

f' = 1
f = x
g = ln(1+x)
g'= 1/(1+x)

Int 1.ln(x+1) = xln(x+1) - Int x/(1+x)
Int 1.ln(x+1) = xln(x+1) - Int 1 - 1/(x+1)
Int 1.ln(x+1) = xln(x+1) - x + ln(x+1) + c
Int 1.ln(x+1) = [x+1]ln(x+1) - x + c