Integral Exercise 11

We are asked to evaluate the following integral:

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

We start by introducing a substitution to simplify the expression:

$$ t = \sqrt{x} $$

Next, we differentiate both sides:

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

Multiplying both sides by 2 gives:

$$ 2 \, dt = \frac{1}{\sqrt{x}} \ dx $$

We now replace $ \frac{1}{\sqrt{x}} \, dx $ with $ 2 \, dt $ in the integral:

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

$$ \int (1 + e^{ \sqrt{x}}) \cdot \frac{1}{\sqrt{x}} \ dx $$

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

Since $ t = \sqrt{x} $, we can rewrite the integral as:

$$ \int (1 + e^t) \cdot 2 \ dt $$

$$ \int (2 + 2e^t) \ dt $$

Splitting the integral gives:

$$ \int 2 \ dt + \int 2e^t \ dt $$

$$ 2t + 2e^t + c $$

Substituting back $ t = \sqrt{x} $, the final result is:

$$ 2 \sqrt{x} + 2 e^{ \sqrt{x} } + c $$

Alternative Method

Let’s now solve the same integral using a slightly different approach:

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

As before, we set:

$$ t = \sqrt{x} $$

Squaring both sides gives:

$$ x = t^2 $$

Now differentiate both sides with respect to $ t $:

$$ dx = 2t \, dt $$

Substitute this into the original integral:

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

Replacing $ x $ with $ t^2 $ gives:

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

Since $ \sqrt{t^2} = t $ for $ t \geq 0 $, we simplify to:

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

$$ \int (1 + e^t) \cdot 2 \, dt $$

$$ \int 2 + 2e^t \ dt $$

Splitting again:

$$ \int 2 \ dt + \int 2e^t \ dt $$

$$ 2t + 2e^t + c $$

Substituting $ t = \sqrt{x} $, we find the same result:

$$ 2 \sqrt{x} + 2 e^{ \sqrt{x} } + c $$

This confirms the solution is correct.

And so on.