Question

Integrate the following with respect to x:

9xe^3x

Answer 1

${\displaystyle \int 9x{\text{e}}^{3x} \text{d}x=9\int x{\text{e}}^{3x} \text{d}x}$

${\displaystyle \int \text{u} \text{dv}=\text{uv}-{\text{u'v}}_1+{\text{u''v}}_2-\text{u}\text{'''}{\text{v}}_3+\text{u}\text{''}{\text{v}}_4-}$   ...........(1)

u = x

u' = 1

u = 0

dv = ${\displaystyle {\text{e}}^{3x}\cdot \text{d}x}$ 

v = ${\displaystyle \int {\text{e}}^{3x}\cdot \text{d}x}$

= ${\displaystyle \frac{{\text{e}}^{3x}}{3}}$

v₁ = ${\displaystyle \int \text{v} \text{d}x}$

= ${\displaystyle \int \frac{{\text{e}}^{3x}}{3}\cdot \text{d}x}$

= ${\displaystyle \frac{1}{3}\times \frac{1}{3} {\text{e}}^{3x}}$ 

v₂ = ${\displaystyle \int {\text{v}}_1 \text{d}x}$

= ${\displaystyle \int \frac{1}{3^2} {\text{e}}^{3x}\cdot \text{d}x}$

= ${\displaystyle \frac{1}{3^2}\times \frac{1}{3} {\text{e}}^{3x}}$

${\displaystyle 9\int x {\text{e}}^{3x} \text{d}x=9[x\frac{{\text{e}}^{3x}}{3}-1\times \frac{1}{3^2} {\text{e}}^{3x}+0\times \frac{1}{3^2} {\text{e}}^{3x}]}$

${\displaystyle 9\int x {\text{e}}^{3x} \text{d}x=9[\frac{x}{3} {\text{e}}^{3x}-\frac{1}{3^2} {\text{e}}^{3x}]}$

${\displaystyle 9\int x {\text{e}}^{3x} \text{d}x=(3x-1){\text{e}}^{3x}+\text{c}}$