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

The integral from 1 to 3 of (x^2+3)^3x is 2560

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integral from 1 to 3 of (x^2+3)^3x

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

2560

Solution steps

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

Apply u-substitution

: \int_{4}^{12} \frac{u ^{3}}{2} d u

= \int_{4}^{12} \frac{u ^{3}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{4}^{12} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{4}^{12}

= \frac{1}{2} [\frac{u ^{4}}{4}]_{4}^{12}

Compute the boundaries:

5120

= \frac{1}{2} \cdot 5120

Simplify

= 2560

\int x x^{2} + 1 d x

\frac{\partial}{\partial x} (ln (2 x + 3 y))

(2, 9), (9, 2)

\int_{0}^{π} 7 x^{2} sin (\frac{1}{2} x) d x

\frac{1}{( 1 - x )}, 0

What is the integral from 1 to 3 of (x^2+3)^3x ?
The integral from 1 to 3 of (x^2+3)^3x is 2560