Integrate (tan^-1 X)/(1+x²) dx
please help, I don't know how to start this one
To integrate the given expression, you can use the substitution method. Let's follow the steps:
Step 1: Choose a substitution variable. In this case, let's choose u = arctan(x).
Step 2: Calculate du/dx by taking the derivative of both sides of our substitution equation:
du/dx = d(arctan(x))/dx = 1/(1 + x^2)
Step 3: Solve the equation dx = du/(1 + u^2) by rearranging it:
dx = du/(1 + arctan^2(x))
Step 4: Replace x and dx in the original integral with the substitution variable u:
Integral [(arctan(x))/(1 + x^2)] dx = Integral [(u)/(1 + ust^2)] du
Step 5: Now the integral is in terms of u, so we can integrate with respect to u.
Integral [(u)/(1 + ust^2)] du = 1/2 * ln(1 + u^2) + C
Step 6: Re-substitute our original variable x in terms of u:
1/2 * ln(1 + u^2) + C = 1/2 * ln(1 + arctan^2(x)) + C
So, the solution to the integral is 1/2 * ln(1 + arctan^2(x)) + C, where C is the constant of integration.