The Antiderivative of: x(arctanx)dx
It's clearly integration by parts with u=arctanx and dv=x, but after that I go down hill. HELP!
To find the antiderivative of the function x(arctanx)dx, you correctly identified that integration by parts is a suitable method to use.
Integration by parts involves applying the formula:
∫u dv = uv - ∫v du
To start, let's assign u = arctanx and dv = x.
Differentiating u with respect to x gives du/dx = 1/(1+x^2).
To solve for dv, let's integrate x with respect to x. Integrating x gives v = (1/2)x^2.
Now we have:
∫x(arctanx)dx = ∫u dv = uv - ∫v du
Plugging in the values we have:
∫x(arctanx)dx = (arctanx)(1/2)x^2 - ∫(1/2)x^2(1/(1+x^2))dx
So far, the integration becomes slightly more complex. However, we can simplify it by using a trigonometric substitution. Let's substitute u = 1+x^2. By differentiating u with respect to x, we have du/dx = 2x, which implies dx = (1/2x) du.
Substituting these values into the equation:
∫x(arctanx)dx = (arctanx)(1/2)x^2 - ∫(1/2)x^2(1/(1+x^2))dx
= (1/2)∫(arctanx)(x^2)dx - (1/4)∫(1/u)du
= (1/2)∫(arctanx)(x^2)dx - (1/4)ln|u| + C
Simplifying further:
∫x(arctanx)dx = (1/2)∫(arctanx)(x^2)dx - (1/4)ln|1+x^2| + C
The antiderivative of x(arctanx)dx is:
∫x(arctanx)dx = (1/2)∫(arctanx)(x^2)dx - (1/4)ln|1+x^2| + C
This is the final result.