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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 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

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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