int frac{a^{sqrt{x}}}{sqrt{x}} dx = | vedclass

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Question

(B) Let $I = \int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx$. Substitute $\sqrt{x} = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx$. Substitute $\sqrt{x} = t$. Differentiating both sides with respect to $x$,we get $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$. Substituting these into the integral: $I = \int a^t (2 dt) = 2 \int a^t dt$. Using the standard integral formula $\int a^t dt = \frac{a^t}{\log_e a} + c$,we get: $I = 2 \left( \frac{a^t}{\log_e a} \right) + c$. Since $\frac{1}{\log_e a} = \log_a e$,we have: $I = 2 a^t \log_a e + c$. Substituting $t = \sqrt{x}$ back,we get: $I = 2 a^{\sqrt{x}} \log_a e + c$.

$\int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx = $

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