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How do you integrate #2 ln (x-5)#?

Answer 1

Luke Alvarez

#color(blue)(int 2 ln(x-5) dx=(2x-10)*ln(x-5)-2x +C)#

The given

#int 2 ln(x-5) dx#

Let #u=ln (x-5)# Let #dv=dx# Let #v=x# Let #du=(1/(x-5))*dx# Using integration by parts

#int u*dv=uv-int v*du#

#int ln(x-5) dx=x*ln(x-5)-int x/(x-5)dx#

#int ln(x-5) dx=x*ln(x-5)-int (x-5+5)/(x-5)dx#

#int ln(x-5) dx=x*ln(x-5)-int (1+5/(x-5))dx#

#int ln(x-5) dx=x*ln(x-5)-x -5*ln(x-5)#

So that

#int 2 ln(x-5) dx=2*[x*ln(x-5)-x -5*ln(x-5)]#

#int 2 ln(x-5) dx=2x*ln(x-5)-2x -10*ln(x-5)+C#

#color(blue)(int 2 ln(x-5) dx=(2x-10)*ln(x-5)-2x +C)#

God bless....I hope the explanation is useful.

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Answer 2

Isaiah Allen

To integrate 2 ln (x - 5), you can use integration by parts. Let u = ln(x - 5) and dv = 2 dx. Differentiate u to find du/dx, and integrate dv to find v. Then, apply the integration by parts formula: ∫u dv = uv - ∫v du. Calculate uv and ∫v du, and subtract the latter from the former to obtain the integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.