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

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UpStudy AI · Accepted Answer

To evaluate the integral

$$
\int \frac{1}{2t^2 + 5} \, dt,
$$

we can follow these steps:

1. **Factor Out the Coefficient of $$ t^2 $$:**

$$
   \int \frac{1}{2t^2 + 5} \, dt = \frac{1}{2} \int \frac{1}{t^2 + \frac{5}{2}} \, dt
   $$

2. **Recognize the Standard Integral Form:**

The integral now resembles the standard form:

$$
   \int \frac{1}{t^2 + a^2} \, dt = \frac{1}{a} \arctan\left(\frac{t}{a}\right) + C
   $$

where $$ a^2 = \frac{5}{2} $$, so $$ a = \frac{\sqrt{10}}{2} $$.

3. **Apply the Standard Formula:**

Substitute $$ a = \frac{\sqrt{10}}{2} $$ into the standard integral:

$$
   \frac{1}{2} \int \frac{1}{t^2 + \left(\frac{\sqrt{10}}{2}\right)^2} \, dt = \frac{1}{2} \cdot \frac{2}{\sqrt{10}} \arctan\left(\frac{2t}{\sqrt{10}}\right) + C
   $$

4. **Simplify the Expression:**

$$
   \frac{1}{\sqrt{10}} \arctan\left(\frac{2t}{\sqrt{10}}\right) + C
   $$

**Final Answer:**

$$
\int \frac{1}{2t^2 + 5} \, dt = \frac{1}{\sqrt{10}} \arctan\left(\frac{2t}{\sqrt{10}}\right) + C
$$
The integral evaluates to $$ \frac{1}{\sqrt{10}} \arctan\left(\frac{2t}{\sqrt{10}}\right) + C $$.

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