Integration by Parts — Algebraic Functions Only

Some products of two algebraic expressions — such as $x(2x+1)^{3}$ — are technically possible to integrate by fully expanding the power and integrating term by term, but that expansion quickly becomes long and error-prone as the power rises. Integration by parts is a technique that avoids the expansion altogether.

🔒 Scope restriction for this WBCHSE Class 12 Commerce syllabus: integration by parts here is used ONLY when BOTH factors of the product are polynomial or algebraic in nature (like $x$ and $(2x+1)^{3}$, or $x$ and $\sqrt{x+1}$). A product of an algebraic factor with a transcendental factor — $x\,e^{x}$, $x\ln x$, $x\sin x$, and similar pairings — is a materially harder application that is explicitly OUTSIDE this chapter's scope; it is never set as a worked example or exercise question here.

The formula

If $u$ and $v$ are both functions of $x$, the product rule for differentiation, $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx}+v\dfrac{du}{dx}$, rearranges (after integrating both sides) into the integration-by-parts formula:

Choosing $u$ and $dv$

For the algebraic-only products in this chapter, the practical rule is: let $u$ be the SIMPLER factor — usually the plain polynomial factor such as $x$ — so that $du$ becomes as simple as possible (often just $dx$), and let $dv$ be the remaining factor, which is then integrated once to find $v$.

Worked pattern: $\displaystyle\int x(2x+1)^{3}\,dx$

Let $u=x$ and $dv=(2x+1)^{3}\,dx$. Then $du=dx$, and integrating $dv$ using standard formula 1 (with the chain rule for the linear inside factor) gives $v = \dfrac{(2x+1)^{4}}{8}$. Substituting into the formula: …