Q.Evaluate $\displaystyle\int x(2x+1)^{3}\,dx$ using integration by parts.

Both factors here — $x$ and $(2x+1)^{3}$ — are algebraic, so integration by parts applies with $u=x$ chosen as the simpler factor.

Both factors, $x$ and $(2x+1)^{3}$, are algebraic — squarely within this chapter's restricted scope for integration by parts. Choose $u=x$ (the simpler factor, so $du$ is as simple as possible) and $dv=(2x+1)^{3}dx$:

Apply the formula $\int u\,dv = uv-\int v\,du$:

Evaluate the remaining integral using standard formula 1 with the chain rule: …

Check by differentiation: using the product rule on the first term and the chain rule on both, $\dfrac{d}{dx}\left[\dfrac{x(2x+1)^{4}}{8}\right]=\dfrac{(2x+1)^{4}}{8}+\dfrac{x\cdot4(2x+1)^{3}\cdot2}{8}=\dfrac{(2x+1)^{4}}{8}+x(2x+1)^{3}$, and $\dfrac{d}{dx}\left[-\dfrac{(2x+1)^{5}}{80}\right]=-\dfrac{5(2x+1)^{4}\cdot2}{80}=-\dfrac{(2x+1)^{4}}{8}$. Adding the two: $\dfrac{ …

Choosing $u=(2x+1)^{3}$ instead of $u=x$, which makes $du$ MORE complicated (not simpler) and leads to a harder integral rather than an easier one — the whole point of choosing $u$ well is to simplify $du$. Also, forgetting the chain-rule factor when integrating $(2x+1)^{3}$ or $( …