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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{n ^{2} + 9}

\frac{d}{d x} (- \frac{18}{x ^{3}})

f (x) = \frac{1}{x - 2}, at x = 4

(\frac{19}{2}, - \frac{5}{3}), (- \frac{3}{2}, - \frac{7}{24})

\int \frac{sec ( x ) tan ( x )}{sec ( x ) - 4} d x

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