Question

Evaluate the indefinite integral by using the substitution uequals=y Superscript 4 Baseline plus 4 y squared plus 6y4+4y2+6 to reduce the integral to standard form. Integral from nothing to nothing 12 (y Superscript 4 Baseline plus 4 y squared plus 6 )squared (y cubed plus 2 y )font size decreased by 6 dy

Answer 1

Complete Question

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Answer:

The solution to the  indefinite integral is  $k = (y^4 + 4y^2 +2)^3 + C$

Step-by-step explanation:

From the question we are told that

    The indefinite integral is  

             $\int\limits {12(y^4 +4y^2 + 1 )(y^3 +2y)} \, dy$

Let

       $\int\limits \, dk = \int\limits {12(y^4 +4y^2 + 1 )(y^3 +2y)} \, dy$

given that

            $u = y^4 +4y^2 +1$

Now differentiating with respect to y

           $\frac{du}{dy} = 4y^3 + 8y$

=>       $\frac{du}{dy} = 4(y^3 + 2y)$

=>     $\frac{du}{4} = (y^3 + 2y)dy$

So

     $k =\int\limits {12(y^4 +4y^2 + 1 )(y^3 +2y)} \, dy \equiv \int\limits {12(u^2 * \frac{1}{4} } \, du$

     $k = 3 \int\limits {u^2 } \, du$

    $k = 3\frac{u^3}{3} + C$

     $k =u^3 +C$

Substituting back y for u

    $k = (y^4 + 4y^2 +2)^3 + C$