## anonymous 4 years ago integrate x^33^x dx

1. anonymous

product of x raised to power three and 3 raised to power x

2. anonymous

i assume this is $\int x^33^xdx$

3. anonymous

ya

4. anonymous

it is a set up for integration by parts, but you have to do it 3 times ! will reduce the power on $x^3$ once each time

5. anonymous

i have assumed f(x)=x^3 and g'(x)=3^x

6. anonymous

for example in the first time you will put $u=x^3, dv=3^xdx, du=3x^2,v=\frac{3^x}{\ln(3)}$ and get $uv-\int vdu=\frac{x^33^x}{\ln(3)}-\frac{3}{\ln(3)}\int x^23^xdx$

7. anonymous

then lather, rinse, repeat. hold on

8. anonymous

easier to read it , click "show steps" http://www.wolframalpha.com/input/?i=x^3*3^x+dx

9. anonymous

how would u integrate 3x^2.3^x/log3

10. anonymous

first of all pull out the constants like i did above and get $-\frac{3}{\ln(3)}\int x^23^xdx$ then repeat the parts, this time with $u=x^2$

11. anonymous

ur ans went over my head

12. anonymous

ok it is confusing, but if you got the first one right you should have a good idea what is going on each time you integrate by parts, you will reduce the power by one when you take the derivative of $x^3, x^2, x$

13. anonymous

don't forget that $\ln(3)$ is a number (constant) so it come right out front of the integral sign

14. anonymous

your next job is really $\int x^2 3^xdx$ the constant just sits out front. integrate by parts again

15. anonymous

let me show u the first step......3x^2.3x/log3-integrate3x^2.3^x/log3 dx

16. anonymous

ah first step is wrong that is the problem

17. anonymous

it is $\int udv=uv-\int vdu$

18. anonymous

i dnt use that formula

19. anonymous

$u=x^3,du = 3x^2, dv = 3^x, v = \int 3^x=\frac{3^x}{\ln(3)}$

20. anonymous

$uv=\frac{x^33^x}{\ln(3)}$

21. anonymous

and i stress again, even though the second one is $-\int \frac{3x^23^x}{\ln(3)}dx$ you should rewrite it immediately as $-\frac{3}{\ln(3)}\int x^23^xdx$

22. anonymous

i m having trouble with the formula implementation......wht to do?