Answer
$$\frac{t^{2}-25}{25 t}+C$$
Work Step by Step
Given $$\int \frac{d t}{t^{2} \sqrt{t^{2}-25}}$$
Let $$ t=5 \sec \theta,\ \ \ \ dt = 5\sec \theta \tan \theta d \theta $$
\begin{aligned}
\int \frac{d t}{t^{2} \sqrt{t^{2}-25}} &=\int \frac{5 \sec \theta \tan \theta d \theta}{(5 \sec \theta)^{2} \sqrt{(5 \sec \theta)^{2}-25}} \\
&=\int \frac{5 \sec \theta \tan \theta d \theta}{25 \sec ^{2} \theta \sqrt{25 \sec ^{2} \theta-25}} \\
&=\int \frac{\tan \theta d \theta}{5 \sec \theta \cdot 5 \sqrt{\sec ^{2} \theta-1}} \\
&=\frac{1}{25} \int \frac{\tan \theta d \theta}{\sec \theta} \\
&=\frac{1}{25} \int \frac{1}{\sec \theta} d \theta \\
&=\frac{1}{25} \int \cos \theta d \theta \\
&=\frac{1}{25} \sin \theta+C\\
&=\frac{t^{2}-25}{25 t}+C
\end{aligned}
