Home > Calculus > Integrals of Rational Functions

What is the antiderivative of #(x+2)/(x+1)#?

Answer 1

Emilia Aguilar

#x+lnabs(x+1)+C#

Using substitution:

Set #u=x+1#. This means that #x=u-1# and #du=dx#.

#int(x+2)/(x+1)dx=int((u-1)+2)/udu=int(u+1)/udu#

#=int1+1/udu=u+lnu+C=x+1+lnabs(x+1)+C#

The constant #1# is absorbed into the #C#, so the antiderivative is

#x+lnabs(x+1)+C#

Using another more intuitive approach:

#int(x+2)/(x+1)dx=int(x+1+1)/(x+1)dx=int(x+1)/(x+1)+1/(x+1)dx#

#=int1+1/(x+1)dx#

Again, substitution can be applied or you could realize that the derivative of #x+1# is #1#, so #int1/(x+1)dx# is #lnabs(x+1)# and #intdx=x#.

Instead of using the method of splitting #x+2# into #x+1+1#, polynomial long division confirms that #(x+2)/(x+1)=1+1/(x+1)#. Polynomial long division is always a good approach when the degree of the numerator equals the degree of the denominator.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

William Abdalla

The antiderivative of (\frac{x+2}{x+1}) is (x + 2 \ln|x+1| + C), where (C) is the constant of integration.

By signing up, you agree to our Terms of Service and Privacy Policy

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.