Q.Integrate the following with respect to $x$:

Part (i): $9xe^{3x}$. Take $u=x,\,dv=e^{3x}dx\Rightarrow du=dx,\,v=\dfrac{e^{3x}}3$.

$\displaystyle\int xe^{3x}dx=\dfrac{xe^{3x}}3-\int\dfrac{e^{3x}}3dx=\dfrac{xe^{3x}}3-\dfrac{e^{3x}}9+c$.

Multiplying by $9$: $9\displaystyle\int xe^{3x}dx=3xe^{3x}-e^{3x}+c$.

Check: $\dfrac d{dx}[3xe^{3x}-e^{3x}]=3e^{3x}+9xe^{3x}-3e^{3x}=9xe^{3x}$ ✓.

Part (ii): $x\sin3x$. Take $u=x,\,dv=\sin3x\,dx\Rightarrow du=dx,\,v=-\dfrac{\cos3x}3$.

$\displaystyle\int x\sin3x\,dx=-\dfrac{x\cos3x}3+\dfrac13\int\cos3x\,dx=-\dfrac{x\cos3x}3+\dfrac{\sin3x}9+c$.

Check: $\dfrac d{dx}\left[-\dfrac{x\cos3x}3+\dfrac{\sin3x}9\right]=-\dfrac{\cos3x}3+x\sin3x+\dfrac{\cos3x}3=x\sin3x$ ✓.

Part (iii): $25xe^{-5x}$. Take $u=x,\,dv=e^{-5x}dx\Rightarrow du=dx,\,v=-\dfrac{e^{-5x}}5$.

$\displaystyle\int xe^{-5x}dx=-\dfrac{xe^{-5x}}5+\dfrac15\int e^{-5x}dx=-\dfrac{xe^{-5x}}5-\dfrac{e^{-5x}}{25}+c$.

Multiplying by $25$: $25\displaystyle\int xe^{-5x}dx=-5xe^{-5x}-e^{-5x}+c$.

Check: $\dfrac d{dx}[-5xe^{-5x}-e^{-5x}]=-5e^{-5x}+25xe^{-5x}+5e^{-5x}=25xe^{-5x}$ ✓.

Part (iv): $x\sec x\tan x$. Since $\dfrac d{dx}\sec x=\sec x\tan x$, take $u=x,\,dv=\sec x\tan x\,dx\Rightarrow du=dx,\,v=\sec x$.

$\displaystyle\int x\sec x\tan x\,dx=x\sec x-\int\sec x\,dx=x\sec x-\log|\sec x+\tan x|+c$.

Check: $\dfrac d{dx}[x\sec x-\log|\sec x+\tan x|]=\sec x+x\sec x\tan x-\sec x=x\sec x\tan x$ ✓.

• Choosing $u=e^{3x}$ (or $\sec x\tan x$) instead of $u=x$, which never terminates
• Forgetting to multiply the whole antiderivative by the leading constant (9 or 25)