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

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

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

integral

x (x^{2} + 4)^{3}

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

Solution steps

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

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

2 x^{- 5}

\frac{( x + 2 )}{( x ^{2} + 2 )}

\frac{( z + 4 )}{( z ^{2} + 2 z + 5 )}

\frac{( 1 - cos ( x ) )}{x}

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