What is the integral of (x^4+3x^2+1)^8(4x^3+6x) ?

The integral of (x^4+3x^2+1)^8(4x^3+6x) is ((x^4+3x^2+1)^9)/9+C

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integral of (x^4+3x^2+1)^8(4x^3+6x)

\int (x^{4} + 3 x^{2} + 1)^{8} (4 x^{3} + 6 x) d x

\frac{( x ^{4} + 3 x ^{2} + 1 ) ^{9}}{9} + C

Solution steps

\int (x^{4} + 3 x^{2} + 1)^{8} (4 x^{3} + 6 x) d x

Apply u-substitution

: \int u^{8} d u

= \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= \frac{u ^{9}}{9}

Substitute back

u = x^{4} + 3 x^{2} + 1

= \frac{( x ^{4} + 3 x ^{2} + 1 ) ^{9}}{9}

Add a constant to the solution

= \frac{( x ^{4} + 3 x ^{2} + 1 ) ^{9}}{9} + C

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What is the integral of (x^4+3x^2+1)^8(4x^3+6x) ?
The integral of (x^4+3x^2+1)^8(4x^3+6x) is ((x^4+3x^2+1)^9)/9+C