How do you integrate by parts: $\int x e^{3} x d x$ $int x e^3xdx$ ?

Question

How do you integrate by parts: $\int x e^{3} x d x$ $int x e^3xdx$ ?

1 Answer

Impact of this question

12212 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-09T07:07:17

This seems to be:

$\int x e^{3 x} d x$ $int xe^(3x)dx$

$e^{3} (x)$ $e^3(x)$ wouldn't make sense ( $e$ $e$ is a constant, not a function), and $e^{3} x \neq {(e x)}^{3}$ $e^3x ne (ex)^3$ (improper implications from $t r i g^{n} (x) = {(t r i g x)}^{n}$ $trig^n(x) = (trigx)^n$ ). $e^{x^{3}}$ $e^(x^3)$ would be very advanced to integrate, and would not be remotely easy by integration by parts. $x^{2} e^{3}$ $x^2e^3$ would be way too simple.

Assuming so...

Let:
$u = x$ $u = x$
$d u = 1 d x$ $du = 1dx$
$d v = e^{3 x} d x$ $dv = e^(3x)dx$
$v = \frac{1}{3} e^{3 x}$ $v = 1/3e^(3x)$

$= u v - \int v d u$ $= uv - intvdu$

$= \frac{x}{3} e^{3 x} - \frac{1}{3} \int e^{3 x} d x$ $= x/3e^(3x) - 1/3inte^(3x)dx$

$= \frac{x}{3} e^{3 x} - \frac{1}{3} [\frac{1}{3} e^{3 x}] + C]$ $= x/3e^(3x) - 1/3[1/3e^(3x)] + C]$

$= \frac{x}{3} e^{3 x} - \frac{1}{9} e^{3 x} + C$ $= x/3e^(3x) - 1/9e^(3x) + C$

or

$= \frac{e^{3 x}}{9} (3 x - 1) + C$ $= e^(3x)/9 (3x - 1) + C$

Science

Math

Humanities

... and beyond

How do you integrate by parts: $\int x e^{3} x d x$ $int x e^3xdx$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate by parts: ∫xe3xdxint x e^3xdx?

1 Answer

Related questions

Impact of this question

How do you integrate by parts: $\int x e^{3} x d x$ $int x e^3xdx$ ?