Home > Calculus > Integrals of Rational Functions

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

Answer 1

This is quite easy, as you can simplify the fraction as follows: #(1+x)/(1+x)^2=1/(1+x)# To do this, you have to point out that the domain of the function is #D=(-infty,-1) cup (-1,+infty)#.

Now you can integrate #int (1+x)/(1+x)^2 dx=int 1/(1+x) dx = ln|1+x|+C#

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

Answer 2

Axel Alvarez

To integrate ( \frac{{1+x}}{{(1+ x)^2}} ) with respect to ( x ), you can first simplify the expression by canceling out the common factor in the numerator and denominator, then integrate the resulting expression.

[ \frac{{1+x}}{{(1+ x)^2}} = \frac{{1+x}}{{1+ 2x + x^2}} ]

[ = \frac{{1+x}}{{(1+x)^2}} ]

[ = \frac{1}{{1+x}} ]

Now, integrate ( \frac{1}{{1+x}} ) with respect to ( x ) using the standard integral formula for ( \frac{1}{{1+x}} ), which is ( \ln|1+x| + C ), where ( C ) is the constant of integration.

Therefore, the integral of ( \frac{1+x}{{(1+ x)^2}} ) with respect to ( x ) is ( \ln|1+x| + C ).

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.