The integration by parts formula is popularly written in the following two forms in calculus.

$(1).\,\,$ $\displaystyle \int{u}dv$ $\,=\,$ $uv$ $-$ $\displaystyle \int{v}du$

$(2).\,\,$ $\displaystyle \int{f(x)}.d\big(g(x)\big)$ $\,=\,$ $f(x).g(x)$ $-$ $\displaystyle \int{g(x)}.d\big(f(x)\big)$

Now, let us learn how to derive the integration by parts formula mathematically in calculus. Firstly, let us consider two functions in terms of $x$. So, the functions can be written as $f(x)$ and $g(x)$ in mathematics.

$f{(x)}$ and $g{(x)}$ are two functions in terms of $x$.

According to the product rule of the derivatives, the derivative of the product of two functions can be written in mathematical form as follows.

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

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

Multiply the differential expressions on both sides of the equation by a differential element $dx$ to write the differential equation in differential form.

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

On the right hand side of the equation, the differential element $dx$ can be distributed across the addition of the two terms by the distributive property of multiplication over the addition.

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

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

Let’s focus on simplifying the mathematical expressions on both sides of differential equation.

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

$\implies$ $\dfrac{d{\,\Big(f(x)g(x)\Big)} \times 1}{dx} \times dx$ $\,=\,$ $f(x) \times \dfrac{d{\,g(x)} \times 1}{dx} \times dx$ $+$ $g(x) \times \dfrac{d{\,f(x)} \times 1}{dx} \times dx$

$\implies$ $d{\,\Big(f(x)g(x)\Big)} \times \dfrac{1}{dx} \times dx$ $\,=\,$ $f(x) \times d{\,g(x)} \times \dfrac{1}{dx} \times dx$ $+$ $g(x) \times d{\,f(x)} \times \dfrac{1}{dx} \times dx$

$\implies$ $d{\,\Big(f(x)g(x)\Big)} \times \dfrac{1 \times dx}{dx}$ $\,=\,$ $f(x) \times d{\,g(x)} \times \dfrac{1 \times dx}{dx}$ $+$ $g(x) \times d{\,f(x)} \times \dfrac{1 \times dx}{dx}$

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

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

$\implies$ $d{\,\Big(f(x)g(x)\Big)} \times 1$ $\,=\,$ $f(x) \times d{\,g(x)} \times 1$ $+$ $g(x) \times d{\,f(x)} \times 1$

$\implies$ $d{\,\Big(f(x)g(x)\Big)}$ $\,=\,$ $f(x) \times d{\,g(x)}$ $+$ $g(x) \times d{\,f(x)}$

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

The above differential equation confuses most of the learners. So, it is better to express it in understandable form. So, denote the functions by two variables, which means $u = f(x)$ and $v = g(x)$.

Now, the above differential can be now written in terms of variables $u$ and $v$.

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

Now, integrate the expressions on both sides of the equation to find the integrals of them.

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

The differential equation is now converted as an integral equation. The equation states that the values of both sides of the expressions are equal. So, the integral constants on both sides of the equation are also equal. For that reason, we can neglect the constant of integration in this case.

It is time to find the integrals. Firstly, let us try to find the integration on left hand side of the equation.

$\implies$ $\displaystyle \int{1 \times d(uv)}$ $\,=\,$ $\displaystyle \int{(udv+vdu)}$

The integral of one with respect to $uv$ can be calculated as per the integral one rule.

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

Now, let us concentrate on finding the integral on the right hand side of the equation. According to the addition rule of integrals, the integral of sum of the functions is equal to the sum of their integrals.

$\implies$ $uv$ $\,=\,$ $\displaystyle \int{u}dv$ $+$ $\displaystyle \int{v}du$

Now, move any one of the two terms from right hand side to left hand side of the equation. In this case, the second term on the right hand side of the equation is moved to left hand side of the equation.

$\implies$ $uv$ $-$ $\displaystyle \int{v}du$ $\,=\,$ $\displaystyle \int{u}dv$

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{u}dv$ $\,=\,$ $uv$ $-$ $\displaystyle \int{v}du$

This integral equation is called the integration by parts in calculus. It is also called the partial integration. The integration by parts formula can also be written in terms of x by relacing the values of $u$ and $v$.

$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \int{f(x)}d\big(g(x)\big)$ $\,=\,$ $f(x).g(x)$ $-$ $\displaystyle \int{g(x)}d\big(f(x)\big)$

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