What is the integral of 6x^2(x^3+1)^{19} ?

The integral of 6x^2(x^3+1)^{19} is 1/10 (x^3+1)^{20}+C

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integral of 6x^2(x^3+1)^{19}

\int 6 x^{2} (x^{3} + 1)^{19} d x

\frac{1}{10} (x^{3} + 1)^{20} + C

Solution steps

\int 6 x^{2} (x^{3} + 1)^{19} d x

Take the constant out:

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

= 6 \cdot \int x^{2} (x^{3} + 1)^{19} d x

Apply u-substitution

= 6 \cdot \int \frac{u ^{19}}{3} d u

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int u^{19} d u

Apply the Power Rule

= 6 \cdot \frac{1}{3} \cdot \frac{u ^{20}}{20}

Substitute back

u = x^{3} + 1

= 6 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 1 ) ^{20}}{20}

Simplify

6 \cdot \frac{1}{3} \cdot \frac{( x ^{3} + 1 ) ^{20}}{20} : \frac{1}{10} (x^{3} + 1)^{20}

= \frac{1}{10} (x^{3} + 1)^{20}

Add a constant to the solution

= \frac{1}{10} (x^{3} + 1)^{20} + C

What is the integral of 6x^2(x^3+1)^{19} ?
The integral of 6x^2(x^3+1)^{19} is 1/10 (x^3+1)^{20}+C