Question

Find the integral. $$ \int \frac{1}{4+(x-1)^{2}} d x $$

Short answer

\(\frac{1}{2} \text{arctan}(\frac{x-1}{2}) + C\

Step-by-step solution

Identify the constants for the integral formula

The integral is given as \(\int \frac{1}{4+(x-1)^{2}} d x\). Comparing this with the standard form \(\int \frac{1}{a^{2}+(x-h)^{2}} d x\), it can be concluded that \(a^{2} = 4\) and \(h = 1\). Thus, \(a = 2\) (taking the positive value as \(a\) is a scaling factor) and \(h = 1\).

Substitute into the integral formula

By putting the values of \(a = 2\) and \(h = 1\) into the standard formula for the integral of this kind \(\frac{1}{a} \text{arctan}(\frac{x-h}{a}) + C\), the equation becomes \(\int \frac{1}{4+(x-1)^{2}} d x = \frac{1}{2} \text{arctan}(\frac{x-1}{2}) + C\).

Write the Final Answer

Therefore, the integral of the given function is \(\frac{1}{2} \text{arctan}(\frac{x-1}{2}) + C\). The \(+C\) represents the constant of integration that arises from the indefinite nature of the integral.