What is the integral of 36x^3(3x^4+3)^5 ?

The integral of 36x^3(3x^4+3)^5 is 1/2 (3x^4+3)^6+C

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integral of 36x^3(3x^4+3)^5

\int 36 x^{3} (3 x^{4} + 3)^{5} d x

\frac{1}{2} (3 x^{4} + 3)^{6} + C

Solution steps

\int 36 x^{3} (3 x^{4} + 3)^{5} d x

Take the constant out:

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

= 36 \cdot \int x^{3} (3 x^{4} + 3)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{12} d u

= 36 \cdot \int \frac{u ^{5}}{12} d u

Take the constant out:

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

= 36 \cdot \frac{1}{12} \cdot \int u^{5} d u

Apply the Power Rule

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

= 36 \cdot \frac{1}{12} \cdot \frac{u ^{6}}{6}

Substitute back

u = 3 x^{4} + 3

= 36 \cdot \frac{1}{12} \cdot \frac{( 3 x ^{4} + 3 ) ^{6}}{6}

Simplify

36 \cdot \frac{1}{12} \cdot \frac{( 3 x ^{4} + 3 ) ^{6}}{6} : \frac{1}{2} (3 x^{4} + 3)^{6}

= \frac{1}{2} (3 x^{4} + 3)^{6}

Add a constant to the solution

= \frac{1}{2} (3 x^{4} + 3)^{6} + C

y \to 0 lim (\frac{0}{y})

f (x) = (x^{2} - 3) (x^{3} - 2 x), at x = - 2

\frac{\partial}{\partial x} (x e^{y} + x^{2} cos (y))

\int e^{1.25 x} d x

\frac{21 x ^{2} + 6 x - 6}{2 x x}

What is the integral of 36x^3(3x^4+3)^5 ?
The integral of 36x^3(3x^4+3)^5 is 1/2 (3x^4+3)^6+C