KarlaKalurky 3 years ago integrate:

1. KarlaKalurky

|dw:1352245389039:dw|

2. KarlaKalurky

i really dont have an idea...

3. KarlaKalurky

|dw:1352246169490:dw|

4. KarlaKalurky

@RolyPoly

5. RolyPoly

Neither do I.. :( Let u=x+4 du = dx => x+2 = u-2 ; x = u-4 $\int (\frac{x+2}{x+4})^2 e^x dx = \int (\frac{u-2}{u})^2 e^{u-4}du = \int\frac{u^2-4u+4}{u^2}e^{u-4}du$$=\int (e^{u-4}-\frac{4e^{u-4}}{u}+\frac{4e^{u-4}}{u^2})du$ I'm sure @hartnn has a better idea!!

6. hartnn

can u find derivative of x/(x+4) ??

7. KarlaKalurky

yes...wait...but why x / x+4?

8. hartnn

find its derivative, u'll come to know.....

9. KarlaKalurky

|dw:1352246668426:dw|

10. hartnn

wait i think i made a mistake....

11. hartnn

u'll need integration by parts. but that becomes very messy...

12. hartnn

i just realized that x/x+4 + d/dx (x/x+4) = [(x+2)/(x+4)]^2

13. hartnn

then using this : $$\\ \huge \color{red}{\int e^x[f(x)+f’(x)]dx=e^xf(x)+c} \\$$ i got the integral as (xe^x)/(x+4) + c

14. hartnn

i don't know how to explain that ....

15. RolyPoly

Reverse: $\frac{d}{dx} e^x f(x)+C= e^xf(x)+e^xf'(x)=e^x(f(x)+f'(x))$So, $\int e^x(f(x)+f'(x)) = e^xf(x)+C$ And the question, Consider f(x) = x/x+4 f'(x) = 4/(x+4) f(x) + f'(x) = [(x+2)/(x+4)]^2 So, $\int [\frac{(x+2)}{(x+4)}]^2e^xdx = \int e^x(\frac{x}{x+4}+\frac{4}{(x+4)^2})dx=e^x(\frac{x}{x+4})+C$ Hail @hartnn

16. KarlaKalurky

yes..that's im about to type lol thanks! @hartnn @RolyPoly

17. hartnn

i meant i can't explain how we can think of f(x) as x/(x+4) all other steps are just use of formula.....

18. hartnn

comes with practice, i guess...