Evaluate int sqrt{e^{4x} + e^{2x}} , dx. | vedclass

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(A) Let $I = \int \sqrt{e^{4x} + e^{2x}} \, dx = \int e^x \sqrt{e^{2x} + 1} \, dx$. Substitute $e^x = v$,then $dv = e^x \, dx$. $I = \int \sqrt{v^2 +

Vedclass · Accepted Answer

$\frac{1}{2} e^x \sqrt{e^{2x} + 1} + \frac{1}{2} \sinh^{-1}(e^x) + c$

Vedclass · Answer

(A) Let $I = \int \sqrt{e^{4x} + e^{2x}} \, dx = \int e^x \sqrt{e^{2x} + 1} \, dx$. Substitute $e^x = v$,then $dv = e^x \, dx$. $I = \int \sqrt{v^2 + 1} \, dv$. Using the standard integral formula $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln|x + \sqrt{x^2 + a^2}| + c$. Here $a = 1$,so $I = \frac{v}{2} \sqrt{v^2 + 1} + \frac{1}{2} \ln|v + \sqrt{v^2 + 1}| + c$. Since $\sinh^{-1}(v) = \ln(v + \sqrt{v^2 + 1})$,we have $I = \frac{v}{2} \sqrt{v^2 + 1} + \frac{1}{2} \sinh^{-1}(v) + c$. Substituting $v = e^x$ back,we get $I = \frac{1}{2} e^x \sqrt{e^{2x} + 1} + \frac{1}{2} \sinh^{-1}(e^x) + c$.

Evaluate $\int \sqrt{e^{4x} + e^{2x}} \, dx$.

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