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

Question

Emily Ammerman · Accepted Answer

# int \ 1/((1+tanx)^2+1) \ dx #
# \ \ \ \= 1/5 {ln(tan^2x+2tanx+2) + arctan(tanx+1)} + ln(tan^2x+1) - x} + C# Let: # I= int \ 1/((1+tanx)^2+1) \ dx # Expanding the denominator gives: # I = int \ 1/(1+2tanx+tan^2x+1) \ dx # # \ \ = int \ 1/(tan^2x + 2tanx + 2) \ dx # Because #d/dxtan^2x = sec^2x# and #sec^2x=tan^2x+1# then we can manipulate the integrand as follows:: # I = int \ 1/(tan^2x + 2tanx + 2) \ sec^2x/(1+tan^2x) \ dx # Now we can simplify the integrand by using the substitution: # t = tan x => (dt)/dx = sec^2x # Which gives us: # I = int \ 1/((t^2 + 2t + 2)(1+t^2)) \ dt # Therefore we have transformed the integral into a simpler problem which can be solved by decomposing the new integrand into partial fraction. I will omit the partial fraction decomposition, bt the resulting expansion is: # I = int \ (2t+3)/(5(t^2+2t+2)) - (2t-1)/(5(t^2+1)) \ dt # # \ \ = 1/5 \ int \ (2t+3)/(t^2+2t+2) - (2t-1)/(t^2+1) \ dt \ \ ..... (star)# Consider the first part of the integral #(star)#: # int \ (2t+3)/(t^2+2t+2) \ dt = int \ (2t+2 + 1)/(t^2+2t+2) \ dt # # " " = int \ (2t+2)/(t^2+2t+2) +1/(t^2+2t+2) \ dt# # " " = int \ (2t+2)/(t^2+2t+2) +1/((t+1)^2+1) \ dt# # " " = ln|t^2+2t+2| + arctan(t+1)# # " " = ln(t^2+2t+2) + arctan(t+1)# And for the second part of the integral #(star)#: # int \ (2t-1)/(t^2+1) \ dt = int \ (2t)/(t^2+1) -1/(t^2+1) \ dt # # " " = ln|t^2+1| - arctant # # " " = ln(t^2+1) - arctant # Note that in both cases the modulus signs can be removed as the logarithmic arguments are both positive. Combining our results into #(star)# we get: # I = 1/5 {ln(t^2+2t+2) + arctan(t+1)} + ln(t^2+1) - arctant} + C# Restoring the substitution then gives s: # I = 1/5 {ln(tan^2x+2tanx+2) + arctan(tanx+1)} + ln(tan^2x+1) - arctantanx} + C# # \ \ = 1/5 {ln(tan^2x+2tanx+2) + arctan(tanx+1)} + ln(tan^2x+1) - x} + C#

Isabella Ackerman · Answer

undefined The integral of $ \frac{1}{(1+\tan(x))^2+1} $ with respect to $ x $ is $ \arctan(\tan(x)) + C $, where $ C $ is the constant of integration.