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How do you integrate #int (3x+5)/(x-2)#?

Answer 1

Emma Aldridge

#3x+11ln(abs(x-2))+C#

First simplify the integrand. This can be done by performing long division, or by my preferred method:

#int(3x+5)/(x-2)dx=int(3(x-2)+11)/(x-2)dx#

#color(white)(int(3x+5)/(x-2)dx)=int(3(x-2))/(x-2)dx+int11/(x-2)dx#

#color(white)(int(3x+5)/(x-2)dx)=3intdx+11intdx/(x-2)#

Both of these are simple integration problems. If you're stuck on the second one, try the substitution #u=x-2=>du=dx# and recall that #int(du)/u=ln(absu)+C#.

#color(white)(int(3x+5)/(x-2)dx)=3x+11ln(abs(x-2))+C#

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Answer 2

Ezra Adams

To integrate ( \frac{3x + 5}{x - 2} ) with respect to ( x ), you can use polynomial long division or partial fraction decomposition to simplify the integrand. Once simplified, you can integrate each term separately. If you use partial fraction decomposition, the result would involve natural logarithms and would be expressed as ( 3\ln|x - 2| + 8x - 17 + C ), where ( C ) is the constant of integration.

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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.