What is the integral of (4x^3+1)^412x^2 ?

The integral of (4x^3+1)^412x^2 is 1/5 (4x^3+1)^5+C

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integral of (4x^3+1)^412x^2

\int (4 x^{3} + 1)^{4} 12 x^{2} d x

\frac{1}{5} (4 x^{3} + 1)^{5} + C

Solution steps

\int (4 x^{3} + 1)^{4} \cdot 12 x^{2} d x

Take the constant out:

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

= 12 \cdot \int (4 x^{3} + 1)^{4} x^{2} d x

Apply u-substitution

: \int \frac{( 4 u + 1 ) ^{4}}{3} d u

= 12 \cdot \int \frac{( 4 u + 1 ) ^{4}}{3} d u

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \int (4 u + 1)^{4} d u

Apply u-substitution

: \int \frac{v ^{4}}{4} d v

= 12 \cdot \frac{1}{3} \cdot \int \frac{v ^{4}}{4} d v

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \int v^{4} d v

Apply the Power Rule

: \frac{v ^{5}}{5}

= 12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{v ^{5}}{5}

Substitute back

= 12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 4 x ^{3} + 1 ) ^{5}}{5}

Simplify

12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{( 4 x ^{3} + 1 ) ^{5}}{5} : \frac{1}{5} (4 x^{3} + 1)^{5}

= \frac{1}{5} (4 x^{3} + 1)^{5}

Add a constant to the solution

= \frac{1}{5} (4 x^{3} + 1)^{5} + C

\frac{d}{d x} (e^{- x^{8}})

\frac{\partial}{\partial x} (sin (\frac{x + y}{x - y}))

x^{3} - 12 x^{2} + 27 x, - x^{3} + 12 x^{2} - 27 x

\frac{d}{d x} (9 sin (x) + 9 cos (x))

x^{\frac{x}{3}}

What is the integral of (4x^3+1)^412x^2 ?
The integral of (4x^3+1)^412x^2 is 1/5 (4x^3+1)^5+C