Find Integral of sqrt(e^(9x)) - Incorrect Solution Explained • Physics Forums

7yler · 2011-02-16T19:37:12+00:00

Find the integral. \int\sqrt{e^{9x}} dx I figured that \int\sqrt{e^{9x}} dx is equal to \int(e^{9x/2}) dx so the integral should simply be e^{9x/2}+C Why isn't this correct?

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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Find Integral of sqrt(e^(9x)) - Incorrect Solution Explained

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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