Home > Calculus > Definite and indefinite integrals

What is #int ( 1+x)/(1+x^2)dx#?

Answer 1

Luke Alvarez

The answer is #=arctan(x)+1/2ln(1+x^2)+C#

The integral is

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

#I=I_1+I_2#

#I_1# is a standard integral

#I_1=arctan(x)#

For #I_2#, perform the substitution

#u=1+x^2#, #=>#, #du=2xdx#

Therefore,

#I_2=1/2int(du)/(u)=1/2ln(u)#

#=1/2ln(1+x^2)#

And finally,

#int((1+x)dx)/(1+x^2)=arctan(x)+1/2ln(1+x^2)+C#

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

Answer 2

Charles Arocha

To evaluate the integral (\int \frac{1+x}{1+x^2} , dx), you can use a simple substitution. Let (u = 1 + x^2), then (du = 2x , dx). Rearrange to solve for (dx), getting (dx = \frac{du}{2x}). Now substitute (u) and (dx) into the integral.

After substitution, the integral becomes (\int \frac{1}{u} , du), which integrates to (\ln|u| + C), where (C) is the constant of integration.

Finally, substitute back (u = 1 + x^2) to get the final result: (\ln|1 + x^2| + 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.