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

Part (i): $x\log x$. Take $u=\log x,\,dv=x\,dx\Rightarrow du=\dfrac1x dx,\,v=\dfrac{x^2}2$.

$\displaystyle\int x\log x\,dx=\dfrac{x^2\log x}2-\int\dfrac{x^2}2\cdot\dfrac1x\,dx=\dfrac{x^2\log x}2-\dfrac12\int x\,dx=\dfrac{x^2\log x}2-\dfrac{x^2}4+c$.

Check: $\dfrac d{dx}\left[\dfrac{x^2\log x}2-\dfrac{x^2}4\right]=x\log x+\dfrac x2-\dfrac x2=x\log x$ ✓.

Part (ii): $27x^2e^{3x}$. Bernoulli with $u=x^2\,(u'=2x,\,u''=2)$, $dv=e^{3x}dx\,(v=\tfrac{e^{3x}}3,\,v_1=\tfrac{e^{3x}}9,\,v_2=\tfrac{e^{3x}}{27})$:

$\displaystyle\int x^2e^{3x}dx=x^2\cdot\dfrac{e^{3x}}3-2x\cdot\dfrac{e^{3x}}9+2\cdot\dfrac{e^{3x}}{27}+c=\dfrac{x^2e^{3x}}3-\dfrac{2xe^{3x}}9+\dfrac{2e^{3x}}{27}+c$.

Multiplying by $27$: $27\displaystyle\int x^2e^{3x}dx=9x^2e^{3x}-6xe^{3x}+2e^{3x}+c$.

Check: $\dfrac d{dx}[9x^2e^{3x}-6xe^{3x}+2e^{3x}]=18xe^{3x}+27x^2e^{3x}-6e^{3x}-18xe^{3x}+6e^{3x}=27x^2e^{3x}$ ✓.

Part (iii): $x^2\cos x$. Bernoulli with $u=x^2\,(u'=2x,\,u''=2)$, $dv=\cos x\,dx\,(v=\sin x,\,v_1=-\cos x,\,v_2=-\sin x)$:

$\displaystyle\int x^2\cos x\,dx=x^2\sin x-2x(-\cos x)+2(-\sin x)+c=x^2\sin x+2x\cos x-2\sin x+c$.

Check: $\dfrac d{dx}[x^2\sin x+2x\cos x-2\sin x]=2x\sin x+x^2\cos x+2\cos x-2x\sin x-2\cos x=x^2\cos x$ ✓. …

Integration by parts — non-integrable $u=\log x$, and Bernoulli …