Chapter 6.2, Problem 4E

To determine

To find: the given indefinite integral.

Answer to Problem 4E

  ${\displaystyle \int \frac{dt}{t^2+1}}={\tan }^{-1}\left(t\right)+C$ .

Explanation of Solution

Given:

  ${\displaystyle \int \frac{dt}{t^2+1}}$ .

Formula used:

  ${\displaystyle \int \frac{dx}{x^2+a^2}}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C$

Calculation:

Consider the given integral,

  $\begin{array}{c}{\displaystyle \int \frac{dt}{t^2+1}}={\displaystyle \int \frac{dt}{t^2+1^2}}\\ =\frac{1}{1}\cdot {\tan }^{-1}\left(\frac{t}{1}\right)+C\\ ={\tan }^{-1}\left(t\right)+C\end{array}$