Intergration problem

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Intergration problem

Calculus1
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i have prblem with the "e"
if it was only x it would be easyier

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Other answers:

You know what \(\dfrac{\mathrm d}{\mathrm dx}~e^{2x}\) is, right?
yes
To find integral of \(5e^{2x}\), you want to look for value \(c\) in \(c~e^{2x}\) such that taking derivaitve it would give you \(5e^{2x}\)
Since \(\dfrac{\mathrm d}{\mathrm dx}~e^{2x} = 2e^{2x}\), you have \(\dfrac{\mathrm d}{\mathrm dx}~c~e^{2x} = 2c~e^{2x}\). Here, we have \(2c = 5\)
i thought we have to take the antider wouldnt it be 5/2e^2x?
So that c is 5/2 Therefore integral of \(5e^{2x}\) is \(\dfrac{5e^{2x}}{2}\)
yes what do i do after ? for substition
substitution? Well, you basically have \(\left.\dfrac{5}{2}e^{2x}~\right|_1^8 = \left(\dfrac{5}{2}e^{2(8)}\right)-\left(\dfrac{5}{2}e^{2(1)}\right)\)
ok
You know how to evaluate \(\displaystyle \int_1^8 5e^{2x}~\mathrm dx?\), right? do same for other terms.
for 4/x i get 0 ?
i am right ?
-1+1 =0
\[\int\dfrac{4}{x}\mathrm~dx = 4\ln(8) - 4\ln(1)\]
idk i might be right , but my final solution is s/2e^14 + 196607.25
damn i am wrong
196607.25 isn't exact value?
k after simplity everything i get 5/2e^2x - 4lnx-3/4x^3/4(x)
k after simplity everything i get 5/2e^2x - 4lnx+3/4x^3/4(x
\[\int\limits \left(\sqrt[3]{x}+5 e^{2 x}-\frac{4}{x}\right) \, dx=\frac{3 x^{4/3}}{4}+\frac{5 e^{2 x}}{2}-4 \log (x) \]\[\frac{\partial \left(\frac{3 x^{4/3}}{4}+\frac{5 e^{2 x}}{2}-4 \log (x)\right)}{\partial x}=\sqrt[3]{x}+5 e^{2 x}-\frac{4}{x} \]
Sounds good

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