int frac{5^{x}}{sqrt{5^{-2x}-5^{2x}}} dx= | vedclass

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Question

(B) Let $I = \int \frac{5^{x}}{\sqrt{5^{-2x}-5^{2x}}} dx$. We can rewrite the denominator as: $\sqrt{5^{-2x}-5^{2x}} = \sqrt{\frac{1}{5^{2x}} - 5^{2x}

Vedclass · Accepted Answer

$\frac{\sin ^{-1}\left(5^{2 x}\right)}{\log 25}+c$

Vedclass · Answer

(B) Let $I = \int \frac{5^{x}}{\sqrt{5^{-2x}-5^{2x}}} dx$. We can rewrite the denominator as: $\sqrt{5^{-2x}-5^{2x}} = \sqrt{\frac{1}{5^{2x}} - 5^{2x}} = \sqrt{\frac{1 - (5^{2x})^2}{5^{2x}}} = \frac{\sqrt{1 - (5^{2x})^2}}{5^x}$. Substituting this into the integral: $I = \int \frac{5^x}{\frac{\sqrt{1 - (5^{2x})^2}}{5^x}} dx = \int \frac{5^{2x}}{\sqrt{1 - (5^{2x})^2}} dx$. Let $t = 5^{2x}$. Then $dt = 5^{2x} \cdot \ln(5) \cdot 2 dx = 2 \ln(5) \cdot 5^{2x} dx$. So,$5^{2x} dx = \frac{dt}{2 \ln(5)} = \frac{dt}{\ln(25)}$. Substituting into the integral: $I = \int \frac{1}{\sqrt{1 - t^2}} \cdot \frac{dt}{\ln(25)} = \frac{1}{\ln(25)} \int \frac{dt}{\sqrt{1 - t^2}}$. $I = \frac{1}{\ln(25)} \sin^{-1}(t) + c = \frac{\sin^{-1}(5^{2x})}{\ln(25)} + c$.

$\int \frac{5^{x}}{\sqrt{5^{-2x}-5^{2x}}} dx=$

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