What is the integral of (x^2-4)^3x ?

The integral of (x^2-4)^3x is 1/8 (x^2-4)^4+C

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integral of (x^2-4)^3x

\int (x^{2} - 4)^{3} x d x

\frac{1}{8} (x^{2} - 4)^{4} + C

Solution steps

\int (x^{2} - 4)^{3} x d x

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = x^{2} - 4

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 4 ) ^{4}}{4} : \frac{1}{8} (x^{2} - 4)^{4}

= \frac{1}{8} (x^{2} - 4)^{4}

Add a constant to the solution

= \frac{1}{8} (x^{2} - 4)^{4} + C

n = 1 \sum \infty \frac{- 4 ^{n}}{4 ^{n}}

\int_{0}^{\infty} e^{- 0.8 x} d x

y^{^{''}} + 8 y^{^{'}} + 32 y = 32, y (0) = 2, y^{^{'}} (0) = - 20

\int cos (x) cos (2 x) cos (3 x) d x

\int_{0}^{\frac{3}{2} 3} \frac{x ^{3}}{9 - x ^{2}} d x

What is the integral of (x^2-4)^3x ?
The integral of (x^2-4)^3x is 1/8 (x^2-4)^4+C