What is the integral of 3x^2(1+x^3)^5 ?

The integral of 3x^2(1+x^3)^5 is 1/6 (1+x^3)^6+C

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integral of 3x^2(1+x^3)^5

\int 3 x^{2} (1 + x^{3})^{5} d x

\frac{1}{6} (1 + x^{3})^{6} + C

Solution steps

\int 3 x^{2} (1 + x^{3})^{5} d x

Take the constant out:

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

= 3 \cdot \int x^{2} (1 + x^{3})^{5} d x

Apply u-substitution

= 3 \cdot \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

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

Substitute back

u = 1 + x^{3}

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of 3x^2(1+x^3)^5 ?
The integral of 3x^2(1+x^3)^5 is 1/6 (1+x^3)^6+C