Integration by parts

From the product rule, backwards. For differentiable $u(x)$ and $v(x)$, the product rule states $\dfrac{d}{dx}[uv] = u'v + uv'$. Integrate both sides with respect to $x$ and rearrange, and the differentiation formula flips into an integration formula:

where: $u$ is the factor you choose to differentiate (it should get simpler under $d/dx$), $v'$ is the factor you choose to integrate (it should have an easy antiderivative $v$), and $u'$ is the derivative of $u$. In the notation NCERT prefers, calling the two factors $u$ (first function) and $v$ (second function, easy to integrate), the same formula reads:

Both versions say the same thing: pick a $u$, pick a $v'$, and the product's integral becomes (something already known) minus (a hopefully-simpler integral).

Choosing $u$ and $v$: the ILATE rule. The single biggest source of marks lost on this topic is picking $u$ and $v$ the wrong way round, which turns a shrinking problem into a growing one. The standard heuristic ranks function types by priority for being chosen as $u$:

Whichever type sits earlier in this list becomes $u$ (the one you differentiate); the other becomes $v'$ (the one you integrate). The logic behind the order is that differentiating an inverse-trig or log function tends to simplify it dramatically (often down to an algebraic fraction), while differentiating a polynomial repeatedly reduces its degree toward zero — both moves shrink the remaining integral. Differentiating an exponential or a plain trig function, by contrast, just reproduces a similar function, so those are better left to be integrated.

A useful shortcut. When the integrand has the special shape $e^x\big[f(x) + f'(x)\big]$, the answer collapses without any real work:

This is a one-line consequence of applying integration by parts to $\int e^x f(x)\,dx$ and watching the leftover $\int e^x f'(x)\,dx$ term cancel against the extra $f'(x)$ term already present.

Products that never simplify away. For integrands like $e^x\sin x$, neither factor is reduced to nothing by repeated differentiation, so the formula must be applied twice. On the second pass, the original integral reappears on the right side, and it can then be solved for algebraically, as in the worked example below.

Functions in disguise. $\int \ln x\,dx$ and $\int \tan^{-1}x\,dx$ do not look like products at all, but they are — with the second factor being the constant function $1$. Writing them as $\int(\ln x)\cdot 1\,dx$ or $\int(\tan^{-1}x)\cdot 1\,dx$ makes $u = \ln x$ (or $\tan^{-1}x$) and $v' = 1$ explicit, and this trick is the only way logarithmic and inverse-trig functions get integrated by this method, since they have no natural "easy to integrate" partner.

Evaluate $\displaystyle\int x\,e^x\,dx$, then extend to the repeated case $\displaystyle\int e^x\sin x\,dx$.

Step 1. By ILATE, "Algebraic" ($x$) outranks "Exponential" ($e^x$), so take $u = x$ and $v' = e^x$, giving $v = \int e^x\,dx = e^x$.

Step 2. Substitute into $\int u\,v'\,dx = uv - \int u'v\,dx$: $\int x e^x\,dx = x\cdot e^x - \int (1)\cdot e^x\,dx = xe^x - e^x + C$. Checking by differentiation, $\dfrac{d}{dx}(xe^x - e^x) = (e^x + xe^x) - e^x = xe^x$, matching the integrand. So $\int x e^x\,dx = xe^x - e^x + C$.

Step 3. Now the harder case, $I = \int e^x\sin x\,dx$, where neither factor disappears. Take $u = \sin x$, $v' = e^x \Rightarrow v = e^x$: $I = e^x\sin x - \int e^x\cos x\,dx$. Apply the formula again to $\int e^x\cos x\,dx$, keeping the same rule (differentiate the trig factor each time): with $u = \cos x$, $v' = e^x \Rightarrow v = e^x$, $\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx = e^x\cos x + I$.

Step 4. Substitute back: $I = e^x\sin x - (e^x\cos x + I) = e^x\sin x - e^x\cos x - I$, so $2I = e^x(\sin x - \cos x)$. Final answer: $I = \displaystyle\int e^x\sin x\,dx = \dfrac{e^x(\sin x - \cos x)}{2} + C$.

• Picking $u$ and $v$ against ILATE — this can make $\int u'v\,dx$ harder than the original integral instead of simpler.
• Adding the constant of integration $C$ after each partial step instead of only once, at the very end.
• Switching which factor is $u$ midway through a repeated application (example: differentiating $e^x$ on the second pass instead of the trig factor) — the original integral then fails to reappear correctly.
• Ignoring the $e^x[f(x)+f'(x)]$ shortcut and grinding through the full formula when it is not needed.
• Not noticing that a "single" function like $\ln x$ or $\tan^{-1}x$ is secretly a product with $1$, and trying to look it up as a standard integral instead.

Ask the tutor: unsure how to pick $u$ and $v$ for a tricky product, or want a walkthrough of a "twice around" integration by parts with a different function pair? Ask the AI tutor for guided practice.