Integration splits neatly over addition and subtraction, so it feels as though it should split over multiplication in the same way. It does not. Integrating each factor of a product on its own and multiplying the results is the single most common mistake in integration.

Note Addition and subtraction really do split

Integration passes term by term through a sum or a difference:

\int \big(f(x) \pm g(x)\big)\,dx = \int f(x)\,dx \;\pm\; \int g(x)\,dx

Take $2x$ , whose integral is $x^{2} + C$ . Break it up across a plus sign and integrate the pieces:

2x = x + x

, so

\displaystyle\int 2x\,dx = \int x\,dx + \int x\,dx = \dfrac{x^{2}}{2} + \dfrac{x^{2}}{2}

⟹ $x^{2} + C$

Break it across a minus sign instead and the pieces still reassemble:

2x = 3x - x

, so

\displaystyle\int 2x\,dx = \int 3x\,dx - \int x\,dx = \dfrac{3x^{2}}{2} - \dfrac{x^{2}}{2}

⟹ $x^{2} + C$

Split the function however you like across $+$ and $-$ , integrate each part, and the answer always comes back correct.

Note Multiplication does not split

There is no matching rule for a product. Writing one down is the mistake:

\int f(x)\,g(x)\,dx \;\neq\; \left( \int f(x)\,dx \right)\!\left( \int g(x)\,dx \right)

Example Watch the wrong method fail

Integrate $\displaystyle\int x^{2}\cdot x\,dx$ , first the wrong way, then the right way.

Integrate each factor separately and multiply the results:

\displaystyle\int x^{2}\,dx = \dfrac{x^{3}}{3}, \qquad \int x\,dx = \dfrac{x^{2}}{2}

⟹ $\dfrac{x^{3}}{3}\cdot\dfrac{x^{2}}{2} = \dfrac{x^{5}}{6}$ — wrong

Simplify the product first, then integrate once:

x^{2}\cdot x = x^{3}

, so

\displaystyle\int x^{3}\,dx = \dfrac{x^{4}}{4} + C

⟹ $\dfrac{x^{4}}{4} + C$ — correct

Differentiating settles which is which:

\dfrac{d}{dx}\!\left(\dfrac{x^{4}}{4}\right) = x^{3}

but

\dfrac{d}{dx}\!\left(\dfrac{x^{5}}{6}\right) = \dfrac{5x^{4}}{6} \neq x^{3}

The two answers are not a small slip apart — they are different functions, and only $\dfrac{x^{4}}{4} + C$ differentiates back to $x^{3}$ .

Note What to do with a product instead

If the product simplifies, simplify it first and integrate once. Collecting $x^{2}\cdot x$ into $x^{3}$ is the whole job; the power rule then finishes it in one step.

If it will not simplify — a genuine function times a function, such as $\displaystyle\int x\,e^{x}\,dx$ — there is a dedicated technique called integration by parts, covered in a later lesson. Multiplying two separate integrals is never the way out.

Summary

Integration distributes over $+$ and $-$ , never over $\times$ .
$\displaystyle\int \big(f \pm g\big)\,dx = \int f\,dx \pm \int g\,dx$ ; there is no product version of this.
Integrating each factor and multiplying gives the wrong answer: $\displaystyle\int x^{2}\cdot x\,dx = \dfrac{x^{4}}{4} + C$ , not $\dfrac{x^{5}}{6}$ .
If the product simplifies, simplify first and integrate once.
If it does not simplify, use integration by parts — a later lesson.

The Common Mistake in Integrating a Product