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How do you integrate #int (1+x) /(1+ x)^2 dx#?

Answer 1

Emilia Aguilar

It is

#int (1+x) /(1+ x)^2 dx=int 1/(1+x)dx=log(x+1)+c#

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Answer 2

Aurora Alvarado

To integrate ( \frac{1+x}{(1+x)^2} ) with respect to ( x ), you can rewrite the expression as ( \frac{1}{1+x} ) and then integrate using the power rule for integration.

The integral of ( \frac{1}{1+x} ) with respect to ( x ) is ( \ln|1+x| + 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.

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