Verified answer

we can use a simple method called tabular formula how?

just find which is easy to differentiate here x^2 or e^2x

of course it will be x^2 because as we differentiate x^2 many times it will reach zero at the end

but e^2x won't

so as a result we use a method called tabular formula

this method works that u differentiate the term which it will reach zero at the end and other one integrate it

line a line b

1- (x^2)-4 e^2x

2- 2x 1/2*e^2x

3- 2 1/4*e^2x

4- 0 1/8*e^2x

this method tell us to take the first term in the line (the one we differentiate) times the integrated part in the second line

so it will be (x^2-4)*1/2*(e^2x)

and same goes for the next lines but for the even lines we add a negative sign for example the next expression will be

-2x*1/4*(e^2x)

then the third one will be

2*1/8*(e^2x)

then we sum all the expressions and this is the final answer

(x^2-4)*1/2*(e^2x)+(-2x*1/4*(e^2x))+(2*1/8*(e^2x))+C

Represent the integrand as e^x.dsinx: ∫e^x cosx dx=e^x sinx -∫sinx e^x dx +c1 Now we have: -∫sinx e^x dx=∫e^x dcosx=e^x cosx - ∫cosx e^x dx + c2 Substituting the above with later: ∫e^x cosx dx = e^x sinx+e^x cosx - ∫e^x cosx dx +c1+c2 hence: ∫e^x cosx dx = 1/2 e^x (sinx+cosx) + c

(1/2e^(2x))*(x^2 - 4) - 1/2x*e^(2x) + 1/4e^(2x) + C

ʃ(x² - 4)*e^(2x) dx

u = x² - 4

dv = e^(2x) dx

v = e^(2x)/2

du = 2x dx

e^(2x)(x²/2 - 2) - ʃx*e^(2x) dx

u = x

dv = e^(2x) dx

v = e^(2x)/2

du = dx

e^(2x)(x²/2 - 2) - [x*e^(2x)/2 - e^(2x)/4] + C

e^(2x)(x²/2 - 2) - x*e^(2x)/2 + e^(2x)/4 + C

t=2x dt/2=dx

1/2((t-4)e^t)dt

1/2(te^t-4e^t)dt

u=t dv=e^t

du=dt v=e^t

-2e^t+1/2(uv-int(vdu))

-2e^t+1/2(t*e^t-e^t)

1/2(e^(2x)*(2x-1))-2*e^(2x)

vote plz