Complete Question
The complete question is shown on the first uploaded image
Answer:
The solution to the indefinite integral is [tex]k = (y^4 + 4y^2 +2)^3 + C[/tex]
Step-by-step explanation:
From the question we are told that
The indefinite integral is
[tex]\int\limits {12(y^4 +4y^2 + 1 )(y^3 +2y)} \, dy[/tex]
Let
[tex]\int\limits \, dk = \int\limits {12(y^4 +4y^2 + 1 )(y^3 +2y)} \, dy[/tex]
given that
[tex]u = y^4 +4y^2 +1[/tex]
Now differentiating with respect to y
[tex]\frac{du}{dy} = 4y^3 + 8y[/tex]
=> [tex]\frac{du}{dy} = 4(y^3 + 2y)[/tex]
=> [tex]\frac{du}{4} = (y^3 + 2y)dy[/tex]
So
[tex]k =\int\limits {12(y^4 +4y^2 + 1 )(y^3 +2y)} \, dy \equiv \int\limits {12(u^2 * \frac{1}{4} } \, du[/tex]
[tex]k = 3 \int\limits {u^2 } \, du[/tex]
[tex]k = 3\frac{u^3}{3} + C[/tex]
[tex]k =u^3 +C[/tex]
Substituting back y for u
[tex]k = (y^4 + 4y^2 +2)^3 + C[/tex]
