\( \int \frac { 1 } { 2 t ^ { 2 } + 5 } d t \)

To evaluate the integral \( \int \frac{1}{2t^2 + 5} dt \), you can start by rewriting the denominator: First, factor out the constant in the denominator: \[ 2t^2 + 5 = 2 \left( t^2 + \frac{5}{2} \right) \] Now, rewrite the integral as: \[ \int \frac{1}{2(t^2 + \frac{5}{2})} dt = \frac{1}{2} \int \frac{1}{t^2 + \frac{5}{2}} dt \] Next, recognize that \( t^2 + \frac{5}{2} \) can be expressed in a more suitable form for integration. You can utilize the standard integral form: \[ \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \] In this case, \( a^2 = \frac{5}{2} \), which means \( a = \sqrt{\frac{5}{2}} = \frac{\sqrt{10}}{2} \). Therefore: \[ \int \frac{1}{t^2 + \frac{5}{2}} dt = \frac{1}{\frac{\sqrt{10}}{2}} \tan^{-1} \left( \frac{t}{\frac{\sqrt{10}}{2}} \right) + C = \frac{2}{\sqrt{10}} \tan^{-1} \left( \frac{2t}{\sqrt{10}} \right) + C \] Now, substituting back into your equation, we have: \[ \frac{1}{2} \int \frac{1}{t^2 + \frac{5}{2}} dt = \frac{1}{2} \cdot \frac{2}{\sqrt{10}} \tan^{-1} \left( \frac{2t}{\sqrt{10}} \right) + C = \frac{1}{\sqrt{10}} \tan^{-1} \left( \frac{2t}{\sqrt{10}} \right) + C \] Thus, the final result for the integral \( \int \frac{1}{2t^2 + 5} dt \) is: \[ \frac{1}{\sqrt{10}} \tan^{-1} \left( \frac{2t}{\sqrt{10}} \right) + C \]