Question

Find \(f\left( x \right)\).

53. \(f'\left( x \right) = \frac{2}{{\left( {1 + {x^2}} \right)}}\) , \(f\left( 0 \right) = - 1\)

Short answer

The required value is, \(f\left( x \right) = 2\arctan x - 1\).

Step-by-step solution

Find anti-derivative

Given that \(f'\left( x \right) = \frac{2}{{\left( {1 + {x^2}} \right)}}\).

The general anti-derivative of the function is \(f\left( x \right) = 2\arctan x + C\).

Find the constant

Now, \(f\left( x \right) = 2\arctan x + C\)

And \(f\left( 0 \right) = - 1\)

Then,

\(\begin{aligned}{c}2\arctan 0 + C &= - 1\\2\left( 0 \right) + C &= - 1\\C &= - 1\end{aligned}\)

Hence \(f\left( x \right) = 2\arctan x - 1\).