Question

How do you integrate \[\int {\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}dx} \]?

Answer 1

Hint: Expand the numerator of the Integrant with the use of an appropriate algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] to simplify and integrate with the use of basic integration identities like \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\], where \[n\] in any number other than \[ - 1\] and for \[n = - 1\].

Complete step by step solution:
Write the given integration.
\[\int {\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}dx} \]

For the given integration, the integrand function is \[\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}\].

Expand the numerator of the integrant function with the use of algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] as shown below.
\[ \Rightarrow \dfrac{{{{\left( {4{x^2}} \right)}^2} + {1^2} - 2\left( {4{x^2}} \right)\left( 1 \right)}}{{{x^3}}}\]

Simplify each term in the numerator as shown below.
\[ \Rightarrow \dfrac{{{4^2}{x^4} + 1 - 8{x^2}}}{{{x^3}}}\]
\[ \Rightarrow \dfrac{{16{x^4} + 1 - 8{x^2}}}{{{x^3}}}\]
\[ \Rightarrow \dfrac{{16{x^4} - 8{x^2} + 1}}{{{x^3}}}\]

Now, simplify the expression by dividing the denominator to each term of numerator as shown below.
\[ \Rightarrow \dfrac{{16{x^4}}}{{{x^3}}} - \dfrac{{8{x^2}}}{{{x^3}}} + \dfrac{1}{{{x^3}}}\]
\[ \Rightarrow 16x - \dfrac{8}{x} + \dfrac{1}{{{x^3}}}\]

Therefore, it is observed that we can write the integrand function is \[\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}\] as \[16x - \dfrac{8}{x} + \dfrac{1}{{{x^3}}}\].
\[\therefore \int {\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}dx} = \int {\left( {16x - \dfrac{8}{x} + \dfrac{1}{{{x^3}}}} \right)dx} \]

Now solve the integration as follow:
\[\begin{array}{c}\int {\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}dx} = \int {\left( {16x - \dfrac{8}{x} + \dfrac{1}{{{x^3}}}} \right)dx} \\ = \int {16xdx} - \int {\dfrac{8}{x}dx} + \int {\dfrac{1}{{{x^3}}}dx} \\ = 16\int {xdx} - 8\int {\dfrac{1}{x}dx} + \int {\dfrac{1}{{{x^3}}}dx} \end{array}\]

Now use the integration identity \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C\], where \[n\] in any number other than \[ - 1\] and for \[n = - 1\], the integration identity is \[\int {{x^{ - 1}}dx} = \ln x + C\].

\[\begin{array}{c}16\int {xdx} - 8\int {\dfrac{1}{x}dx} + \int {\dfrac{1}{{{x^3}}}dx} = 16\left[ {\dfrac{{{x^2}}}{2}} \right] - 8\left[ {\ln x} \right] + \left[ {\dfrac{{{x^{ - 2}}}}{{ - 2}}} \right] + C\\ = 18{x^2} - 8\ln x - \dfrac{1}{{2{x^2}}} + C\end{array}\]

Therefore, the integration of the given expression \[\dfrac{{{{\left( {4{x^2} - 1} \right)}^2}}}{{{x^3}}}\] with respect to \[x\] is \[18{x^2} - 8\ln x - \dfrac{1}{{2{x^2}}} + C\]. Here, \[C\] represents the arbitrary constant.

Note: Integration are of two kinds, Indefinite integrals and Definite integrals. Indefinite integrals are the one with no limits as given in this question, here we have to add an arbitrary constant \[C\,\] in the end to show that it is not a particular solution for the given function.
Definite integrals are the one with limits under which the solution of definite integral is applicable, so there is no need to add an arbitrary constant \[C\,\] in the end. Integration is the inverse operation of differentiation but there is no product law or quotient law in integration therefore we expand any complex expression in a simpler expression and use the knowledge of differential formulas of the same type of expressions to obtain the integration.