What is the integral of 8(y^4+4y^2+1)^2(y^3+2y) ?

The integral of 8(y^4+4y^2+1)^2(y^3+2y) is 2/3 (y^4+4y^2+1)^3+C

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integral of 8(y^4+4y^2+1)^2(y^3+2y)

\int 8 (y^{4} + 4 y^{2} + 1)^{2} (y^{3} + 2 y) d y

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

Solution steps

\int 8 (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

= 8 \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

= 8 \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

= 8 \cdot \frac{1}{4} \cdot \int u^{2} d u

Apply the Power Rule

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

= 8 \cdot \frac{1}{4} \cdot \frac{u ^{3}}{3}

Substitute back

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

= 8 \cdot \frac{1}{4} \cdot \frac{( y ^{4} + 4 y ^{2} + 1 ) ^{3}}{3}

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} + 3 (1 + x^{2}) y = 0

\frac{d}{d m} (mi)

a e^{v} + \frac{b}{v} + \frac{c}{v ^{2}}

\frac{x}{6 x + 4}

\frac{d}{d x} (3 csc (x) + \frac{5}{x})

What is the integral of 8(y^4+4y^2+1)^2(y^3+2y) ?
The integral of 8(y^4+4y^2+1)^2(y^3+2y) is 2/3 (y^4+4y^2+1)^3+C