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

The integral of x^2(x^3+3)^6 is 1/21 (x^3+3)^7+C

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

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

\frac{1}{21} (x^{3} + 3)^{7} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ( u ^{\frac{3}{2}} + 3 ) ^{6}}{2 x} d u

= \int \frac{( u + 3 ) ^{6}}{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 + 3)^{6} d u

Apply u-substitution

: \int v^{6} d v

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

Apply the Power Rule

: \frac{v ^{7}}{7}

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

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 3 ) ^{7}}{7} : \frac{1}{21} (x^{3} + 3)^{7}

= \frac{1}{21} (x^{3} + 3)^{7}

Add a constant to the solution

= \frac{1}{21} (x^{3} + 3)^{7} + C

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What is the integral of x^2(x^3+3)^6 ?
The integral of x^2(x^3+3)^6 is 1/21 (x^3+3)^7+C