General Power Rule of Integration

General Power Rule of integration is another important formula of integration. And this rules need derivative of the given function within the problem.

The general power rule of integration is of the form

\int {{{\left[ {f\left( x \right)} \right]}^n}f'\left( x \right)dx = } \frac{{{{\left[ {f\left( x \right)} \right]}^{n + 1}}}}{{n + 1}} + c

Now consider

\begin{gathered} \frac{d}{{dx}}\left[ {\frac{{{{\left( {f\left( x \right)} \right)}^{n + 1}}}}{{n + 1}} + c} \right] = \frac{d}{{dx}}\left[ {\frac{{{{\left( {f\left( x \right)} \right)}^{n + 1}}}}{{n + 1}}} \right] + \frac{d}{{dx}}\left( c \right) \\ \Rightarrow \frac{d}{{dx}}\left[ {\frac{{{{\left( {f\left( x \right)} \right)}^{n + 1}}}}{{n + 1}} + c} \right] = \left( {n + 1} \right)\frac{{{{\left[ {f\left( x \right)} \right]}^{n + 1 - 1}}}}{{n + 1}}f'\left( x \right) + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {\frac{{{{\left( {f\left( x \right)} \right)}^{n + 1}}}}{{n + 1}} + c} \right] = {\left[ {f\left( x \right)} \right]^n}f'\left( x \right) \\ \Rightarrow {\left[ {f\left( x \right)} \right]^n}f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{{\left( {f\left( x \right)} \right)}^{n + 1}}}}{{n + 1}} + c} \right]\,\,\,\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right) \\ \end{gathered}

Integrating both sides of equation (i) with respect to x, we have

 \Rightarrow \int {{{\left[ {f\left( x \right)} \right]}^n}f'\left( x \right)} = \int {\frac{d}{{dx}}\left[ {\frac{{{{\left( {f\left( x \right)} \right)}^{n + 1}}}}{{n + 1}} + c} \right]\,} \,\,\,\,\,\,{\text{ - - - }}\left( {\text{i}} \right)

Since integration and differentiation are reverse processes to each other , so the integral sign \int {} and \frac{d}{{dx}} on the right side will cancel each other, i.e.

 \Rightarrow \int {{{\left[ {f\left( x \right)} \right]}^n}f'\left( x \right)} \,dx = \frac{{{{\left[ {f\left( x \right)} \right]}^{n + 1}}}}{{n + 1}} + c

Example: Evaluate the integral \int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx} with respect to x

We have integral

I = \int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx}

Here f\left( x \right) = 1 + \frac{1}{{{x^2}}} = 1 + {x^{ - 2}} implies that f'\left( x \right) = 1 - 2{x^{ - 3}} = - \frac{2}{{{x^3}}}

Multiplying and dividing by  - 2 the given integral, we have

\int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx} = - \frac{1}{2}\int { - \frac{2}{{{x^3}}}} {\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{5}{3}}}dx

Using general power formula of integration, we have

\begin{gathered} \int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx} = - \frac{1}{2}\frac{1}{{\frac{5}{3} + 1}}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{5}{3} + 1}} + c \\ \Rightarrow \int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx} = \frac{3}{{16}}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{8}{3}}} + c \\ \end{gathered}