What is the integral of (x^2+3)^32x ?

The integral of (x^2+3)^32x is 1/4 (x^2+3)^4+C

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integral of (x^2+3)^32x

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ^{3}}{2} d u

= 2 \cdot \int \frac{u ^{3}}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{4}}{4}

Substitute back

u = x^{2} + 3

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{5} y^{5})

\int \frac{v + 3}{v - 1} d v

\frac{d}{d x} (3 (5 - x)^{2})

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

f (x) = (4 t - 1) (3 t - 6)^{- 1}

What is the integral of (x^2+3)^32x ?
The integral of (x^2+3)^32x is 1/4 (x^2+3)^4+C