If int frac{e^x}{sqrt{e^{2x}+4e^x+13}} dx = l | vedclass

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(B) Let $I = \int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx$. Substitute $u = e^x$,then $du = e^x dx$. The integral becomes $I = \int \frac{du}{\sqrt{u^2+4

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(B) Let $I = \int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx$. Substitute $u = e^x$,then $du = e^x dx$. The integral becomes $I = \int \frac{du}{\sqrt{u^2+4u+13}}$. Complete the square for the quadratic expression: $u^2+4u+13 = (u+2)^2 + 9 = (u+2)^2 + 3^2$. Using the standard integral formula $\int \frac{dx}{\sqrt{x^2+a^2}} = \log |x + \sqrt{x^2+a^2}| + c$,we get: $I = \log |(u+2) + \sqrt{(u+2)^2 + 3^2}| + c$. Substituting $u = e^x$ back,we get: $I = \log |e^x + 2 + \sqrt{e^{2x}+4e^x+13}| + c$. Comparing this with the given expression $\log |e^{ax}+2+\sqrt{e^{2x}+4e^x+13}| + c$,we find that $a = 1$.

If $\int \frac{e^x}{\sqrt{e^{2x}+4e^x+13}} dx = \log \left|e^x+2+\sqrt{e^{2x}+4e^x+13}\right|+c$,(where $c$ is the constant of integration),then the value of $a$ in the expression $\log \left|e^{ax}+2+\sqrt{e^{2x}+4e^x+13}\right|+c$ is equal to

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