# The General Power Rule of Integration

The general power rule of integration is another important formula of integration, and this rule needs th 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, the integral sign $\int {}$ and $\frac{d}{{dx}}$ on the right side will cancel each other out, 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 the given integral by $– 2$, 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 the general power formula of integration, we have
$\begin{gathered} \int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx} = – \frac{1}{2}\frac{1}{{\frac{5}{3} + 1}}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{5}{3} + 1}} + c \\ \Rightarrow \int {\frac{1}{{{x^3}}}{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{5}{3}}}dx} = \frac{3}{{16}}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{8}{3}}} + c \\ \end{gathered}$