I have to use integration by parts to integrate tan^-1 (1/x)dx. I'm not sure what to use as u or dv. We were taught to use uv-integral of (vdu) to do this.
look for a u that does not get too messy on differentiation, and a v that can be integrated without too much troubls
here, it seems logical to use u = arctan(1/x) since we know we're going to get rid of the nasty arctan, and the chain rule will toss in a 1/x^2:
u = arctan(1/x)
du = 1/(1/x^2 + 1)* (-1/x^2) dx
= -1/(1+x^2) dx
dv = dx
v = x
uv-Int v*du = x*arctan(1/x) + Int(x/(1+x^2)) dx
Int v*du = 1/2 * ln(1+x^2)
So, the final answer is x*arctan(1/x) + 1/2 ln(1+x^2)
go ahead -- take the derivative and watch things cancel out!!
To solve this problem using integration by parts, we will identify the functions u and dv.
Given the integral ∫tan^(-1)(1/x) dx, we can rewrite it as ∫tan^(-1)(x^(-1)) dx to make the expression clearer.
Now, let's identify the functions u and dv in the formula for integration by parts:
u = tan^(-1)(x^(-1))
dv = dx
To determine which function to choose as u and which one as dv, we follow a suggested order: logarithmic functions, inverse trigonometric functions, algebraic functions, and trigonometric functions.
In this case, tan^(-1)(x^(-1)) is an inverse trigonometric function, while dx is simply a differential. According to the suggested order, we should choose u as tan^(-1)(x^(-1)) and dv as dx.
Now, let's differentiate u and find its corresponding du:
du = (1/1+(1/x^2)) * ((-1/x^2) * (1)) dx
= -dx/(x^2 + 1)
Next, integrate dv to obtain v:
∫dv = x
Now we can apply the integration by parts formula:
∫tan^(-1)(x^(-1)) dx = uv - ∫v du
Substituting the values for u, v, and du into the formula:
∫tan^(-1)(x^(-1)) dx = tan^(-1)(x^(-1)) * x - ∫x * (-dx/(x^2 + 1))
Simplifying further:
∫tan^(-1)(x^(-1)) dx = x*tan^(-1)(x^(-1)) + ∫dx/(x^2 + 1)
Now, we have a relatively easier integral to evaluate ∫dx/(x^2 + 1), which can be solved using a basic trigonometric substitution.
Let's solve ∫dx/(x^2 + 1) briefly as it is a well-known integral:
Using the substitution x = tanθ, we have dx = sec^2θdθ, and x^2 + 1 = tan^2θ + 1 = sec^2θ.
∫dx/(x^2 + 1)
= ∫sec^2θdθ/(sec^2θ)
= ∫dθ
= θ + C
Substituting back, we have ∫dx/(x^2 + 1) = θ + C.
Returning to the previous equation:
∫tan^(-1)(x^(-1)) dx = x*tan^(-1)(x^(-1)) + θ + C
Finally, replace θ with tan^(-1)(x):
∫tan^(-1)(x^(-1)) dx = x*tan^(-1)(x^(-1)) + tan^(-1)(x) + C
So, the integral of tan^(-1)(1/x) dx is x*tan^(-1)(x^(-1)) + tan^(-1)(x) + C, where C represents the constant of integration.