What is the integral of x^2(6x^3+4)^5 ?

The integral of x^2(6x^3+4)^5 is 1/108 (6x^3+4)^6+C

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integral of x^2(6x^3+4)^5

\int x^{2} (6 x^{3} + 4)^{5} d x

\frac{1}{108} (6 x^{3} + 4)^{6} + C

Solution steps

\int x^{2} (6 x^{3} + 4)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{18} d u

= \int \frac{u ^{5}}{18} d u

Take the constant out:

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

= \frac{1}{18} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{1}{18} \cdot \frac{u ^{6}}{6}

Substitute back

u = 6 x^{3} + 4

= \frac{1}{18} \cdot \frac{( 6 x ^{3} + 4 ) ^{6}}{6}

Simplify

\frac{1}{18} \cdot \frac{( 6 x ^{3} + 4 ) ^{6}}{6} : \frac{1}{108} (6 x^{3} + 4)^{6}

= \frac{1}{108} (6 x^{3} + 4)^{6}

Add a constant to the solution

= \frac{1}{108} (6 x^{3} + 4)^{6} + C

\int - 4 x^{6} + 6 x^{4} + 3 x^{3} + 6 d x

\int 6 x x - 5 d x

\int (5 x + 1) d x

\int 5 e^{- 5 t} d t

\int \frac{cos ^{3} ( t )}{sin ( t )} d t

What is the integral of x^2(6x^3+4)^5 ?
The integral of x^2(6x^3+4)^5 is 1/108 (6x^3+4)^6+C