Proof of Integration by parts

The mathematical formula for the integration by parts can be derived in integral calculus by the concepts of differential calculus.

Product Rule of Differentiation

$f{(x)}$ and $g{(x)}$ are two functions in terms of $x$. As per the product rule of differentiation, the derivative of the product of the functions $f{(x)}$ and $g{(x)}$ can be written into its equivalent form.

$\dfrac{d}{dx}{\, \Big({f{(x)}}{g{(x)}}\Big)}$ $\,=\,$ $f{(x)}\dfrac{d}{dx}{\, {g{(x)}}}$ $+$ $g{(x)}\dfrac{d}{dx}{\, {f{(x)}}}$

Product Rule in Differential form

Take $u = f{(x)}$ and $v = g{(x)}$. Now, express the derivative product rule in differential form.

$\implies$ $\dfrac{d}{dx}{\, (uv)}$ $\,=\,$ $u\dfrac{d}{dx}{\, v}$ $+$ $v\dfrac{d}{dx}{\, u}$

$\implies$ $\dfrac{d(uv)}{dx}$ $\,=\,$ $u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}$

Multiply both sides of the equation by the differential element $dx$.

$\implies$ $\dfrac{d(uv)}{dx} \times dx$ $\,=\,$ $\Big(u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}\Big) \times dx$

$\implies$ $\dfrac{d(uv)}{dx} \times dx$ $\,=\,$ $u\dfrac{dv}{dx} \times dx$ $+$ $v\dfrac{du}{dx} \times dx$

$\implies$ $d(uv) \times \dfrac{dx}{dx}$ $\,=\,$ $udv \times \dfrac{dx}{dx}$ $+$ $vdu \times \dfrac{dx}{dx}$

$\implies$ $d(uv) \times 1$ $\,=\,$ $udv \times 1$ $+$ $vdu \times 1$

$\implies$ $d(uv)$ $\,=\,$ $udv$ $+$ $vdu$

Integrate the Differential equation

Now, integrate both sides of the differential equation.

$\implies$ $\displaystyle \int d(uv)$ $\,=\,$ $\displaystyle \int udv$ $+$ $\displaystyle \int vdu$

$\implies$ $uv$ $\,=\,$ $\displaystyle \int udv$ $+$ $\displaystyle \int vdu$

$\implies$ $uv$ $-$ $\displaystyle \int vdu$ $\,=\,$ $\displaystyle \int udv$

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int udv$ $\,=\,$ $uv$ $-$ $\displaystyle \int vdu$