Integration by parts is a formula for transforming the integral of the multiplication of two functions. It may be possible to calculate complex integrals by transforming. It is effective for several patterns, especially it is used when the differentiation of one function becomes a simple form or for integration involving the form \(\log x\).

Example :

(Case 1) Differentiation of one becomes easier.

Example : Find \(\int x\sin x\ dx\).

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Solution: Since the derivative of \(x\) is \(1\) (that is \(\frac{d}{dx} x=1\)), we set \(f (x) =x\).

Suppose we choose \(f(x)=x\) and \(g'(x)=\sin x\) (since \(f'(x)\) becomes a simple function). Then, we \(f'(x)=1\) and \(g(x)=-\cos x\). (We can choose any antiderivative of \(g'(x)\).) Thus, using the formula, we have

$$\begin{eqnarray*}\int x\sin x\ dx&=&f(x)g(x)-\int f'(x)g(x) \ dx\\&=&x\cdot (-\cos x)-\int 1\cdot (-\cos x)\ dx\\&=&-x\cos x+\int \cos x\ dx\\&=&-x\cos x+\sin x+C\end{eqnarray*}$$

(Case 2) Forms involving \(\log x\).

Example : Find \(\int \ln x\ dx\)

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Solution: Since the derivative of \(\ln x\) is \(\frac{1}{x}\) (that is \(\frac{d}{dx}\ln x=\frac{1}{x}\)), we set \(f (x) = \ln x\).

Suppose we choose \(f(x)=\ln x\) and \(g'(x)=1\). Then, \(f'(x)=\frac{1}{x}\) and \(g(x)=x\). Thus, we have

$$\begin{eqnarray*}\int \ln x\ dx&=&f(x)g(x)-\int f'(x)g(x) \ dx\\&=&(\ln x)\cdot x-\int \frac{1}{x}\cdot x\ dx\\&=&x\ln x-\int 1\ dx\\&=&x\ln x-x+C\end{eqnarray*}$$

Proof

The product Rule states that if \(f\)and \(g\) are differentiable functions, then

$$\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+ g'(x)f(x)$$

In the notation for indefinite integrals, this equation becomes

$$f(x)g(x)+C=\int f'(x)g(x)\ dx+\int g'(x)f(x)\ dx$$

Therefore, we can arrange the equation as below:

$$\int f(x)g'(x)\ dx=f(x)g(x)-\int f'(x)g(x) \ dx$$

Notice that we can omit the constant of integration C since both sides contains indefinite integrals.