What is the integral of 12(y^4+4y^2+1)^2(y^3+2y) ?

The integral of 12(y^4+4y^2+1)^2(y^3+2y) is (y^4+4y^2+1)^3+C

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integral of 12(y^4+4y^2+1)^2(y^3+2y)

\int 12 (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

(y^{4} + 4 y^{2} + 1)^{3} + C

Solution steps

\int 12 (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \int (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

Apply u-substitution

: \int \frac{u ^{2}}{4} d u

= 12 \cdot \int \frac{u ^{2}}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 12 \cdot \frac{1}{4} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 12 \cdot \frac{1}{4} \cdot \frac{u ^{3}}{3}

Substitute back

u = y^{4} + 4 y^{2} + 1

= 12 \cdot \frac{1}{4} \cdot \frac{( y ^{4} + 4 y ^{2} + 1 ) ^{3}}{3}

Simplify

12 \cdot \frac{1}{4} \cdot \frac{( y ^{4} + 4 y ^{2} + 1 ) ^{3}}{3} : (y^{4} + 4 y^{2} + 1)^{3}

= (y^{4} + 4 y^{2} + 1)^{3}

Add a constant to the solution

= (y^{4} + 4 y^{2} + 1)^{3} + C

\frac{x}{x + 3}

\int 6 x + 1 d x

\frac{e ^{t} - cos ( t )}{t}

\int - 2 x^{- 1} \cdot x^{- 2} d x

y^{^{''}} = - k^{2} \cdot y

What is the integral of 12(y^4+4y^2+1)^2(y^3+2y) ?
The integral of 12(y^4+4y^2+1)^2(y^3+2y) is (y^4+4y^2+1)^3+C