int(dx)/(sqrtxe^((40sqrtx)))=

Make Substitution: Simplify the integral by making a substitution. Let u = sqrt(x), which implies that x = u^2. Now, differentiate both sides with respect to u to find dx in terms of du. (d)/(du)(u^2) = 2u dx = 2u duRewrite in Terms of u: Rewrite the integral in terms of u. The integral becomes: int(dx)/(sqrt(x)e^(40sqrt(x))) = int(2u du)/(u e^(40u)) Simplify the integrand by canceling u in the numerator and denominator. int(2u du)/(u e^(40u)) = int(2 du)/(e^(40u))Evaluate Integral: Evaluate the integral. The integral of 2/e^(40u) with respect to u is a standard exponential integral. int(2 du)/(e^(40u)) = -2/(40) * 1/e^(40u) + C = -1/(20) * 1/e^(40u) + CSubstitute Back: Substitute back the original variable x. Since u = sqrt(x), we have: -1/(20) * 1/e^(40u) + C = -1/(20) * 1/e^(40sqrt(x)) + CWrite Final Answer: Write the final answer. The integral of (dx)/(sqrt(x)e^(40sqrt(x))) is: -1/(20) * 1/e^(40sqrt(x)) + C

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$\int \frac{d x}{\sqrt{x} e^{(40 \sqrt{x})}}=$

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Evaluate definite integrals using the power rule

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\int \frac{d x}{\sqrt{x} e^{(40 \sqrt{x})}}=

Make Substitution: Simplify the integral by making a substitution. $\newline$ Let $u = \sqrt{x}$ , which implies that $x = u^2$ . Now, differentiate both sides with respect to $u$ to find $dx$ in terms of $du$ . $\newline$ $\frac{d}{du}(u^2) = 2u$ $\newline$ $dx = 2u \, du$
Rewrite in Terms of $u$ : Rewrite the integral in terms of $u$ . The integral becomes: $\int\frac{\mathrm{d}x}{\sqrt{x}e^{40\sqrt{x}}} = \int\frac{2u \mathrm{d}u}{u e^{40u}}$ Simplify the integrand by canceling $u$ in the numerator and denominator. $\int\frac{2u \mathrm{d}u}{u e^{40u}} = \int\frac{2 \mathrm{d}u}{e^{40u}}$
Evaluate Integral: Evaluate the integral. $\newline$ The integral of $\frac{2}{e^{40u}}$ with respect to $u$ is a standard exponential integral. $\newline$ $\int \frac{2 \, du}{e^{40u}} = -\frac{2}{40} \cdot \frac{1}{e^{40u}} + C$ $\newline$ $= -\frac{1}{20} \cdot \frac{1}{e^{40u}} + C$
Substitute Back: Substitute back the original variable $x$ . $\newline$ Since $u = \sqrt{x}$ , we have: $\newline$ $-\frac{1}{20} \cdot \frac{1}{e^{40u}} + C = -\frac{1}{20} \cdot \frac{1}{e^{40\sqrt{x}}} + C$
Write Final Answer: Write the final answer. $\newline$ The integral of $\frac{dx}{\sqrt{x}e^{40\sqrt{x}}}$ is: $\newline$ $-\frac{1}{20} \times \frac{1}{e^{40\sqrt{x}}} + C$