What is the integral of (x^3+1)^63x^2 ?

The integral of (x^3+1)^63x^2 is 1/7 (x^3+1)^7+C

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integral of (x^3+1)^63x^2

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ^{6}}{3} d u

= 3 \cdot \int \frac{u ^{6}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u^{6} d u

Apply the Power Rule

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

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{7}}{7}

Substitute back

u = x^{3} + 1

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x^3+1)^63x^2 ?
The integral of (x^3+1)^63x^2 is 1/7 (x^3+1)^7+C