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

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

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

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

\frac{( x ^{3} + 4 x ) ^{3}}{3} + C

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

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

= \frac{u ^{3}}{3}

Substitute back

u = x^{3} + 4 x

= \frac{( x ^{3} + 4 x ) ^{3}}{3}

Add a constant to the solution

= \frac{( x ^{3} + 4 x ) ^{3}}{3} + C

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\frac{d}{d x} (\frac{x ^{2} + 2 x}{( x + 1 ) ^{2}})

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