nml10
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- 2017
How would you integrate (x^5)(e^x^3)?
Integration by Parts (1 Viewer)
- Thread starter nml10
- Start date
ohexploitable
hey
Rinse and repeat.
ohexploitable
hey
Better still,
(xe^{x^2})\,dx\\\\\text{Let }u=x^2,\,du=2x\,dx\\\\=\frac{1}{2}\int u^2 e^u\,du)
Now this should be super duper easy.
Now this should be super duper easy.
hscishard
Active Member
Don't have latex but...
Following how textbooks show it, let u=x^3 and v' = e^(x^3)x^2
Will be simple from now on
Following how textbooks show it, let u=x^3 and v' = e^(x^3)x^2
Will be simple from now on
hscishard
Active Member
Didn't read the question properly lol it's x^3
Rinse and repeat.
i would do this
let
a = x ^ 3
da = 3x^2
hence the integral of
(x^5)(e^x^3)
becomes
(x^3)(x^2)(e^x^3)
u substitute everything in
[a(e^a)]/3
u use int by parts for this take out the 1/3 for this part
let u = a , du = da
v = e^a dv = e^a
so the integral is equal to uv - int(udv)
which is equal to
a(e^a) - int(e^a) = a(e^a) -e^a + C
= [x^3(e^x^3)-(e^x^3) ]/3 + C
let
a = x ^ 3
da = 3x^2
hence the integral of
(x^5)(e^x^3)
becomes
(x^3)(x^2)(e^x^3)
u substitute everything in
[a(e^a)]/3
u use int by parts for this take out the 1/3 for this part
let u = a , du = da
v = e^a dv = e^a
so the integral is equal to uv - int(udv)
which is equal to
a(e^a) - int(e^a) = a(e^a) -e^a + C
= [x^3(e^x^3)-(e^x^3) ]/3 + C