What is the integral of x^2(x^3+7)^{15} ?

The integral of x^2(x^3+7)^{15} is 1/48 (x^3+7)^{16}+C

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integral of x^2(x^3+7)^{15}

\int x^{2} (x^{3} + 7)^{15} d x

\frac{1}{48} (x^{3} + 7)^{16} + C

Solution steps

\int x^{2} (x^{3} + 7)^{15} d x

Apply u-substitution

= \int \frac{u ^{15}}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u^{15} d u

Apply the Power Rule

= \frac{1}{3} \cdot \frac{u ^{16}}{16}

Substitute back

u = x^{3} + 7

= \frac{1}{3} \cdot \frac{( x ^{3} + 7 ) ^{16}}{16}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 7 ) ^{16}}{16} : \frac{1}{48} (x^{3} + 7)^{16}

= \frac{1}{48} (x^{3} + 7)^{16}

Add a constant to the solution

= \frac{1}{48} (x^{3} + 7)^{16} + C

What is the integral of x^2(x^3+7)^{15} ?
The integral of x^2(x^3+7)^{15} is 1/48 (x^3+7)^{16}+C