Integration by parts

The formula. If $u$ and $v$ are differentiable functions of $x$, running the product rule for differentiation backwards and integrating gives

where: $u$ is the factor chosen to be differentiated, $v$ is the factor chosen to be integrated first, $\dfrac{du}{dx}$ is the derivative of $u$, and $\int v\,dx$ is an antiderivative of $v$. In the compact notation you will use most often, writing $dv = v\,dx$, this reads $\int u\,dv = uv - \int v\,du$.

Choosing $u$: the ILATE rule. With two factors in front of you, decide which is $u$ using the priority order Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. Whichever type sits earlier in ILATE becomes $u$, because it usually gets simpler on differentiation, while the later type integrates easily and becomes $dv$. In $\int x\,e^x\,dx$, Algebraic beats Exponential, so $u=x$; differentiating $x$ once kills it, while $e^x$ integrates to itself.

The "multiply by 1" trick. When only one function is visible — $\int \log x\,dx$ or $\int \tan^{-1}x\,dx$ — write it as (function)$\times 1$. Take $u$ as the function and $dv = dx$, so $v = x$. For $\int \log x\,dx$: $u = \log x$, $dv=dx$, giving $du = \tfrac{1}{x}dx$, $v=x$, and

The same move handles $\int \tan^{-1}x\,dx$ and $\int \sin^{-1}x\,dx$: put the inverse-trig function as $u$ against $dv = dx$, then simplify what remains, often with a quick substitution.

A shortcut worth memorising. When the integrand has the exact shape $e^x[f(x)+f'(x)]$, skip integration by parts entirely:

where: $f(x)$ is any differentiable function and $f'(x)$ its derivative — recognising this pattern in the integrand saves a full application of the by-parts formula.

Evaluate $\displaystyle\int x\,e^{x}\,dx$.

Step 1. Choose $u$ and $dv$. By ILATE, Algebraic ($x$) outranks Exponential ($e^x$), so take $u = x$ and $dv = e^x\,dx$.

Step 2. Differentiate $u$ and integrate $dv$: $du = dx$, and $v = e^x$.

Step 3. Substitute into the formula: $\displaystyle\int x e^x\,dx = x\,e^x - \int e^x\,dx = x e^x - e^x + C = e^{x}(x-1) + C$.

Step 4. Check by differentiating the answer: $\dfrac{d}{dx}\big[e^x(x-1)+C\big] = e^x(x-1) + e^x(1) = e^x\cdot x$, which matches the original integrand. Answer: $\displaystyle\int x e^x\,dx = e^{x}(x-1) + C$.

Competitive stretch

Extend the same idea to $\int x^2\sin x\,dx$, which needs parts twice. Take $u=x^2$, $dv=\sin x\,dx$, so $du=2x\,dx$, $v=-\cos x$, giving $\int x^2\sin x\,dx = -x^2\cos x + \int 2x\cos x\,dx$. Applying parts again to $\int 2x\cos x\,dx$ (with $u=2x$, $dv=\cos x\,dx$) gives $2x\sin x + 2\cos x + C$. Combining: $\int x^2\sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + C$. When repeated parts instead brings back a multiple of the original integral — as in $\int e^{ax}\sin bx\,dx$ — treat the result as an algebraic equation in the unknown integral and solve for it, rather than applying parts forever.

• Picking $u$ and $dv$ the wrong way round. Taking $u=e^x$ in $\int xe^x\,dx$ still "works" formally but the resulting integral never simplifies — always check ILATE priority first.
• Dropping the second integral. Students often compute $u\int v\,dx$ and stop, forgetting the $-\int(du/dx)\big(\int v\,dx\big)dx$ term entirely.
• Sign slips when $u$ or $v$ involves sine or cosine — track each term's sign carefully rather than trusting memory.
• Giving up after one application. If the new integral is still a product (as in $\int x^2\sin x\,dx$), it usually needs a second pass of the same rule, not a different method.
• Forgetting to simplify before integrating again. After the first application, the leftover integral $\int(du/dx)\big(\int v\,dx\big)dx$ is often itself a product; tidy the coefficients first so the second pass of the formula is applied to the cleanest possible expression.

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