Integral Exercise 21

We are asked to evaluate the following integral:

$$ \int \frac{e^{\sqrt{2x+1}}}{\sqrt{2x+1}} \ dx $$

To solve it, we use the substitution method, introducing the auxiliary variable \( t = e^{\sqrt{2x+1}} \).

We differentiate \( t \) with respect to \( x \) using the chain rule:

$$ dt = \frac{1}{2\sqrt{2x+1}} \cdot 2 \cdot e^{\sqrt{2x+1}} \ dx $$

which simplifies to:

$$ dt = \frac{e^{\sqrt{2x+1}}}{\sqrt{2x+1}} \ dx $$

Solving for \( dx \), we get:

$$ dx = \frac{\sqrt{2x+1}}{e^{\sqrt{2x+1}}} \ dt $$

Now we substitute this expression for \( dx \) back into the original integral:

$$ \int \frac{e^{\sqrt{2x+1}}}{\sqrt{2x+1}} \cdot \frac{\sqrt{2x+1}}{e^{\sqrt{2x+1}}} \ dt $$

The exponential and square root terms cancel out, leaving:

$$ \int 1 \ dt $$

which integrates immediately to:

$$ t + c $$

Recalling that \( t = e^{\sqrt{2x+1}} \), we substitute back:

$$ \int \frac{e^{\sqrt{2x+1}}}{\sqrt{2x+1}} \ dx = e^{\sqrt{2x+1}} + c $$

This is the final result.