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

To solve the integral \( \int \frac { 1 } { 2 t ^ { 2 } + 5 } d t \), we first recognize that it resembles a standard integral form. We can factor out the constant from the denominator: \[ \int \frac{1}{2(t^2 + \frac{5}{2})} dt = \frac{1}{2} \int \frac{1}{t^2 + \frac{5}{2}} dt. \] Using the formula for the integral of the form \( \int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \), where \( a^2 = \frac{5}{2} \) gives \( a = \sqrt{\frac{5}{2}} \). Thus, the integral becomes: \[ = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{5}{2}}} \tan^{-1}\left(\frac{t}{\sqrt{\frac{5}{2}}}\right) + C. \] Calculating \( \frac{1}{\sqrt{\frac{5}{2}}} \): \[ = \sqrt{\frac{2}{5}}. \] Putting it all together, we get: \[ = \frac{1}{2} \cdot \sqrt{\frac{2}{5}} \tan^{-1}\left(t \sqrt{\frac{2}{5}}\right) + C. \] The final result is: \[ \int \frac{1}{2t^2 + 5} dt = \frac{1}{2} \sqrt{\frac{2}{5}} \tan^{-1}\left(t \sqrt{\frac{2}{5}}\right) + C. \]