Solving for the Antiderivative: 2/x - 5e^5x | Math Homework Help

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Homework Statement


Sorry I have class at 8 and I am getting myself confused...
What is the anti-derivative of 2/x - 5e^5x


Homework Equations





The Attempt at a Solution


So I did this...
2 * 1/x - 5e^5x
Anti-derivative:
2ln - 5e^5x
 
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Is an Anti-Derivitive ...another word for Integration?
if 1/x =lnx ...looks like it.
[Int]2/x - [Int]5e^5x
2[int]1/x - 5[int]e^5x
2Lnx - 5[1/5e^5x]
Ans: 2Lnx -e^5x + k

Am I right?
 

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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