What is the integral of (3x^2+6x)^{51}(x+1) ?

The integral of (3x^2+6x)^{51}(x+1) is 1/312 (3x^2+6x)^{52}+C

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integral of (3x^2+6x)^{51}(x+1)

\int (3 x^{2} + 6 x)^{51} (x + 1) d x

\frac{1}{312} (3 x^{2} + 6 x)^{52} + C

Solution steps

\int (3 x^{2} + 6 x)^{51} (x + 1) d x

Apply u-substitution

: \int \frac{u ^{51}}{6} d u

= \int \frac{u ^{51}}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int u^{51} d u

Apply the Power Rule

: \frac{u ^{52}}{52}

= \frac{1}{6} \cdot \frac{u ^{52}}{52}

Substitute back

u = 3 x^{2} + 6 x

= \frac{1}{6} \cdot \frac{( 3 x ^{2} + 6 x ) ^{52}}{52}

Simplify

\frac{1}{6} \cdot \frac{( 3 x ^{2} + 6 x ) ^{52}}{52} : \frac{1}{312} (3 x^{2} + 6 x)^{52}

= \frac{1}{312} (3 x^{2} + 6 x)^{52}

Add a constant to the solution

= \frac{1}{312} (3 x^{2} + 6 x)^{52} + C

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What is the integral of (3x^2+6x)^{51}(x+1) ?
The integral of (3x^2+6x)^{51}(x+1) is 1/312 (3x^2+6x)^{52}+C