Question

Short answer

Integration by parts method - \(\int {f(x)g'(x)dx = f(x)g(x) - \int {g(x)f'(x)dx} } \)

Or \(\int {udv} = uv - \int {vdu} fg\)

Step-by-step solution

Step-1: Rule for Integration by parts

INTEGRATION BY PARTS – Every differentiation rule functions a corresponding integration rule. For instance, the Substitution Rule for integration is same as Chain-Rule for differentiation. The rule that corresponds to the Product Rule for differentiation is know as the rule for integration by parts.

According to Product Rule, if andare differentiable functions, then

\(\frac{d}{{dx}}\left( {f(x)g(x)} \right) = f(x)g'(x) + g(x)f'(x)\)

for indefinite integrals, this equation will be

\(\int {\frac{d}{{dx}}f(x)g(x)dx} = \int {\left( {f(x)g'(x) + g(x)f'(x)} \right)} dx\)

\( \Rightarrow \)\(f(x)g(x) = \int {\left( {f(x)g'(x) + g(x)f'(x)} \right)} dx\)

\( \Rightarrow \)\(f(x)g(x) = \int {f(x)g'(x)dx + \int {g(x)f'(x)dx} } \)

\( \Rightarrow \)\(\int {f(x)g'(x)dx = f(x)g(x) - \int {g(x)f'(x)dx} } \)

Let \(u = f(x)\) and \(v = g(x)\)

Then \(du = f'(x)dx\) and \(dv = g'(x)dx\)

After putting these values inside the above equation, we get

\(\int {udv} = uv - \int {vdu} \) …. (1)

Step-2: Explanation with example

In practice many times we use it as a substitute rule

Example - \(\int {\ln xdx} \)

Taking \(u = \ln x\) and \(dv = dx\)

Then \(du = \frac{1}{x}dx\) and \(v = x\)

Put these values in equation (1), we get

\(\int {\ln xdx} = x\ln x - \int {x\frac{{dx}}{x}} \)

\( \Rightarrow \)\(\int {\ln xdx} = x\ln x - \int {dx} \)

\( \Rightarrow \)\(\int {\ln xdx} = x\ln x - x + C\)

Step-3: Final Answer

Rule of Integration by parts - \(\int {f(x)g'(x)dx = f(x)g(x) - \int {g(x)f'(x)dx} } \)

Or \(\int {udv} = uv - \int {vdu} \)