Evaluate the integral: K = integral e^(sqrt x) dx

Evaluate the integral: $$ K = \int e^{\sqrt x}\,dx. $$

Explanation of Method

Substituting $u=\sqrt x$ turns $dx$ into $2u\,du$, converting the integral into a polynomial-times-exponential form resolved by a single integration by parts.

Substitute $u=\sqrt x$ and express $dx$ in terms of $du$.
Apply integration by parts to $\int 2ue^u\,du$.
Substitute back and simplify.

Step 1 · Substitution

Let $u=\sqrt x$, so $du=\dfrac{dx}{2\sqrt x}$, giving $dx=2u\,du$. Substituting into the integral:

$$ K = \int e^u\cdot 2u\,du. $$

Step 2 · Integration by parts

Let $p=2u$, $dq=e^u\,du$, so $dp=2\,du$, $q=e^u$:

$$ K = 2ue^u - \int 2e^u\,du. $$

Step 3 · Integrate

Since $\int e^u\,du=e^u$:

$$ K = 2ue^u - 2e^u + C. $$

Step 4 · Substitute back

Substituting back $u=\sqrt x$:

$$ K = 2\sqrt xe^{\sqrt x} - 2e^{\sqrt x} + C. $$

Summary Table

Quantity	Value
Substitution	$u=\sqrt x$
$dx$ in terms of $du$	$2u\,du$
Antiderivative	$2ue^u-2e^u$

Additional Notes

The domain requires $x\geq0$, where $\sqrt x$ is real.
Substituting for the argument of an exponential when that argument is itself a radical is a standard first move — it always turns $\int f(\sqrt x)\,dx$ into $\int 2u\,f(u)\,du$, trading the radical for a factor of $2u$.
The result can be checked by differentiating $2\sqrt xe^{\sqrt x}-2e^{\sqrt x}$ and confirming it returns $e^{\sqrt x}$.

Result

The solution to the integral is:

$$ K = 2\sqrt xe^{\sqrt x} - 2e^{\sqrt x} + C. $$