What is the integral of (x^2+6x)^{24}(x+3) ?

The integral of (x^2+6x)^{24}(x+3) is 1/50 (x^2+6x)^{25}+C

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integral of (x^2+6x)^{24}(x+3)

\int (x^{2} + 6 x)^{24} (x + 3) d x

\frac{1}{50} (x^{2} + 6 x)^{25} + C

Solution steps

\int (x^{2} + 6 x)^{24} (x + 3) d x

Apply u-substitution

= \int \frac{u ^{24}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{24} d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{u ^{25}}{25}

Substitute back

u = x^{2} + 6 x

= \frac{1}{2} \cdot \frac{( x ^{2} + 6 x ) ^{25}}{25}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 6 x ) ^{25}}{25} : \frac{1}{50} (x^{2} + 6 x)^{25}

= \frac{1}{50} (x^{2} + 6 x)^{25}

Add a constant to the solution

= \frac{1}{50} (x^{2} + 6 x)^{25} + C

What is the integral of (x^2+6x)^{24}(x+3) ?
The integral of (x^2+6x)^{24}(x+3) is 1/50 (x^2+6x)^{25}+C