Evaluate the integral: ${\int }_{}^{}\arctan \left(\frac{1}{x}\right)dx$

• A. $x\arctan \left(\frac{1}{x}\right)+\frac{1}{2}\ln \left(x^2+1\right)+C$
• B. $x^2\arctan \left(\frac{1}{x}\right)+C$
• C. $\arctan \left(\frac{1}{x}\right)+C$
• D. $x\arctan \left(x\right)-\frac{1}{2}\ln \left(x^2+1\right)+C$

Answer: (A)

Verified step by step guidance

1. Step 1: Recognize that the integral involves the function arctan(1/x). To solve this, we will use integration by parts, which is based on the formula: ∫u dv = uv - ∫v du.
2. Step 2: Choose u = arctan(1/x) and dv = dx. This means we need to compute du and v. First, differentiate u to find du: u = arctan(1/x) implies du = -1/(x^2 + 1) dx. Next, integrate dv = dx to find v: v = x.
3. Step 3: Substitute into the integration by parts formula: ∫arctan(1/x) dx = uv - ∫v du = x arctan(1/x) - ∫x(-1/(x^2 + 1)) dx.
4. Step 4: Simplify the remaining integral: ∫x(-1/(x^2 + 1)) dx = -∫x/(x^2 + 1) dx. To solve this, use substitution. Let w = x^2 + 1, so dw = 2x dx. Rewrite the integral as -1/2 ∫1/w dw, which simplifies to -1/2 ln|w| + C.
5. Step 5: Substitute back w = x^2 + 1 into the result and combine terms: The final expression becomes x arctan(1/x) + (1/2) ln(x^2 + 1) + C.