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

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

\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

\int (2 x^{4} + 2 x^{3} - x + 7) d x

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\int \frac{1}{t ^{3} + 1} d t

\int \frac{x + x ^{3}}{6 + x ^{2} - x ^{4}} d x

\int \frac{( 4 x ^{2} + 6 )}{x ^{3} + 3 x} d 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