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

The integral of (6x^2+4x)(x^3+x^2+1)^6 is 2/7 (x^3+x^2+1)^7+C

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

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

\frac{2}{7} (x^{3} + x^{2} + 1)^{7} + C

Solution steps

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

Apply u-substitution

: \int 2 u^{6} d u

= \int 2 u^{6} d u

Take the constant out:

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

= 2 \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 2 \cdot \frac{u ^{7}}{7}

Substitute back

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

= 2 \cdot \frac{( x ^{3} + x ^{2} + 1 ) ^{7}}{7}

Simplify

2 \cdot \frac{( x ^{3} + x ^{2} + 1 ) ^{7}}{7} : \frac{2}{7} (x^{3} + x^{2} + 1)^{7}

= \frac{2}{7} (x^{3} + x^{2} + 1)^{7}

Add a constant to the solution

= \frac{2}{7} (x^{3} + x^{2} + 1)^{7} + C

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What is the integral of (6x^2+4x)(x^3+x^2+1)^6 ?
The integral of (6x^2+4x)(x^3+x^2+1)^6 is 2/7 (x^3+x^2+1)^7+C