The Integration-by-Parts Theorem and the LIATE Rule

Theorem. If $u,v$ are differentiable functions of $x$, then $\int uv\,dx=u\int v\,dx-\int\left(\dfrac{du}{dx}\int v\,dx\right)dx$.

Reasoning: let $w=\int v\,dx$, so $v=\frac{dw}{dx}$. Then $\frac{d}{dx}(uw)=u\frac{dw}{dx}+w\frac{du}{dx}=uv+w\frac{du}{dx}$. Integrating both sides and using the definition of integration, $uw=\int uv\,dx+\int w\frac{du}{dx}\,dx$, i.e. $u\int v\,dx=\int uv\,dx+\int\frac{du}{dx}\left(\int v\,dx\right)dx$ — rearranged, this is exactly the stated theorem.

Choosing $u$ — the LIATE rule. The factor taken as $u$ (the one that gets differentiated) should be the one appearing earliest in the order Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential. The wrong choice can make the integral harder rather than easier: for $\int xe^x\,dx$, taking $u=x,\ v=e^x$ (correct LIATE order) gives $I=xe^x-\int e^x\,dx=xe^x-e^x+c$ in one short step. Reversing the choice, $u=e^x,\ v=x$, gives $I=\frac12e^xx^2-\frac12\int e^xx^2\,dx$ — a harder integral than the one we started with. This illustrates why the LIATE order matters. Likewise for $\int x\sin x\,dx$: LIATE puts $x$ (Algebraic) ahead of $\sin x$ (Trigonometric), so $u=x$, $v=\sin x$: $I=x(-\cos x)-\int(1)(-\cos x)\,dx=-x\cos x+\sin x+c$.

Further worked examples, all following the LIATE order:

$\int x^2\,5^x\,dx$. $u=x^2$ (Algebraic beats Exponential), applied twice in succession (since differentiating $x^2$ twice reaches a constant): $I=\dfrac{5^x}{\log5}\left[x^2-\dfrac{2x}{\log5}+\dfrac{2}{(\log5)^2}\right]+c$.

$\int x\tan^{-1}x\,dx$. $u=\tan^{-1}x$ (Inverse trig beats Algebraic): $I=\tan^{-1}x\cdot\dfrac{x^2}2-\displaystyle\int\dfrac{1}{1+x^2}\cdot\dfrac{x^2}2\,dx$. Since $\dfrac{x^2}{1+x^2}=1-\dfrac1{1+x^2}$, the remaining integral is $\frac12\left[x-\tan^{-1}x\right]$. Result: $I=\dfrac12x^2\tan^{-1}x-\dfrac12x+\dfrac12\tan^{-1}x+c$.

$\int\dfrac{x}{1-\sin x}\,dx$. First rationalise: multiply by $\frac{1+\sin x}{1+\sin x}$, giving $\dfrac{x(1+\sin x)}{\cos^2x}=x(\sec^2x+\sec x\tan x)$. By parts with $u=x$ on each piece: $I=x\tan x-\log|\sec x|+x\sec x-\log|\sec x+\tan x|+c$.

$\int e^{2x}\sin3x\,dx$ (cyclic/recurring integration by parts). By parts twice (either factor may be $u$ each time, but keep the same choice throughout): the first application gives $I=-\frac13e^{2x}\cos3x+\frac23\int e^{2x}\cos3x\,dx$; applying by parts again to the remaining integral gives $\int e^{2x}\cos3x\,dx=\frac13e^{2x}\sin3x-\frac23I$. Substituting back: $I=-\frac13e^{2x}\cos3x+\frac23\left[\frac13e^{2x}\sin3x-\frac23I\right]$, i.e. $I+\frac49I=\frac{e^{2x}}9[2\sin3x-3\cos3x]$, so $I=\dfrac{e^{2x}}{13}[2\sin3x-3\cos3x]+c$. (The same recurring trick, worked out once for general $a,b$, gives the standing results $\int e^{ax}\sin(bx+c)\,dx=\frac{e^{ax}}{a^2+b^2}[a\sin(bx+c)-b\cos(bx+c)]+c$ and $\int e^{ax}\cos(bx+c)\,dx=\frac{e^{ax}}{a^2+b^2}[a\cos(bx+c)+b\sin(bx+c)]+c$.) …