Question

Integrate by using the substitution suggested in bracket:
$\int {12\left( {{y^4} + 4{y^2} + 1} \right)\left( {{y^3} + 2y} \right)dy} $, $\left( {use,u = {y^4} + 4{y^2} + 1} \right)$

Answer 1

Hint: First find the differentiation of the term given in the problem, $u = {y^4} + 4{y^2} + 1$ with respect to $y$ and then simplify the result and try to obtain $\left( {{y^3} + 2y} \right)$.
Use these values to make the substitution in the given integral and then find the integral after the substitution. Finally, re-substitute the value of $u$ the obtained solution.
This will give the required result.

Complete step-by-step answer:
Consider the given integral as:
$I = \int {12\left( {{y^4} + 4{y^2} + 1} \right)\left( {{y^3} + 2y} \right)dy} $
The goal of the problem is to find the value of the integral using the substitution given in the problem.
$\left( {use,u = {y^4} + 4{y^2} + 1} \right)$
$u = {y^4} + 4{y^2} + 1$
Differentiate both sides with respect to y.
$\dfrac{{du}}{{dy}} = \dfrac{d}{{dy}}\left( {{y^4} + 4{y^2} + 1} \right)$
$\dfrac{{du}}{{dy}} = 4{y^3} + 8y$
Simplify the result by taking $4$ as a common in the integrand, then we have the result:
$\dfrac{{du}}{{dy}} = 4\left( {{y^3} + 2y} \right)$
$ \Rightarrow \dfrac{{du}}{4} = \left( {{y^3} + 2y} \right)dy$
Substitute $\dfrac{{du}}{4} = \left( {{y^3} + 2y} \right)dy$ and $u = {y^4} + 4{y^2} + 1$ in the integral $I$, so we have
$I = \int {12\left( u \right)\dfrac{{du}}{4}} $
Simplify the obtained result and express again in simplified form:
$I = 3\int {\left( u \right)du} $
Integrate with respect to $u$ and obtain the integral.
$I = 3\left( {\dfrac{{{u^2}}}{2}} \right) + C$, where $C$ is the integral constant.
$I = \dfrac{3}{2}\left( {{u^2}} \right) + C$
Now, substitute the value $u = {y^4} + 4{y^2} + 1$ into the equation:
$I = \dfrac{3}{2}{\left( {{y^4} + 4{y^2} + 1} \right)^2} + C$
The obtained integral of the given problem is $\dfrac{3}{2}{\left( {{y^4} + 4{y^2} + 1} \right)^2} + C$, where $C$ is the integral constant.

Note:The integration by substitution is also said as “The reverse chain rule”.
This is the method to integrate in some special cases. Let $f\left( {g\left( x \right)} \right)$be the integrand and we have to find the integral of the function $\left[ {f\left( {g\left( x \right)} \right)g'\left( x \right)} \right]$, then we use this method of integral.
So, the integral is given as:
$ \Rightarrow \int {\left[ {f\left( {g\left( x \right)} \right)g'\left( x \right)} \right]dx} $
Now, assume that$g\left( x \right) = u$ and differentiate both sides with respect to $x$.
$g'\left( x \right)dx = du$
Now, make the substitution $g\left( x \right) = u$ and $g'\left( x \right)dx = du$in the integral.
$ \Rightarrow \int {f\left( u \right)du} $
Now, we can easily find the integral and re-substitute the value of $u$ in the resultant integral.