Find $f$. \[ f^{\prime}(t)=\frac{12}{1+t^{2}}, \quad f(1)=0 \]
Solution
Step 1 ：Given the derivative function \(f'(t) = \frac{12}{1+t^2}\), we need to find the original function \(f(t)\) by integrating \(f'(t)\) with respect to \(t\).
Step 2 ：The integral of \(f'(t)\) is \(\int f'(t) dt = \int \frac{12}{1+t^2} dt\).
Step 3 ：This is a standard integral, the integral of \(\frac{1}{1+t^2}\) is \(\arctan(t)\), so we have: \(= 12 \int \frac{1}{1+t^2} dt = 12 \arctan(t) + C\), where \(C\) is the constant of integration.
Step 4 ：We are given that \(f(1) = 0\), so we substitute these values into the equation to find \(C\): \(0 = 12 \arctan(1) + C\).
Step 5 ：Since \(\arctan(1) = \frac{\pi}{4}\), we have: \(0 = 12 \cdot \frac{\pi}{4} + C\). Solving for \(C\) gives \(C = -3\pi\).
Step 6 ：So, the function \(f(t)\) is: \(f(t) = 12 \arctan(t) - 3\pi\).
Step 7 ：This is the simplest form of the function, and it satisfies the given condition \(f(1) = 0\). Therefore, the final answer is \(\boxed{f(t) = 12 \arctan(t) - 3\pi}\).
