What is the integral of 2x(x^2+2)^8 ?

The integral of 2x(x^2+2)^8 is 1/9 (x^2+2)^9+C

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integral of 2x(x^2+2)^8

\int 2 x (x^{2} + 2)^{8} d x

\frac{1}{9} (x^{2} + 2)^{9} + C

Solution steps

\int 2 x (x^{2} + 2)^{8} 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)^{8} d x

Apply u-substitution

: \int \frac{u ^{8}}{2} d u

= 2 \cdot \int \frac{u ^{8}}{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^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{9}}{9}

Substitute back

u = x^{2} + 2

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 2 ) ^{9}}{9}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 2 ) ^{9}}{9} : \frac{1}{9} (x^{2} + 2)^{9}

= \frac{1}{9} (x^{2} + 2)^{9}

Add a constant to the solution

= \frac{1}{9} (x^{2} + 2)^{9} + C

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

x = 0, x = 5 - y^{2}

\frac{8}{( s + 2 ) ^{2}}

9 y^{^{''}} + 60 y^{^{'}} + 100 y = 0

8 (x + 2)^{\frac{3}{2}}

What is the integral of 2x(x^2+2)^8 ?
The integral of 2x(x^2+2)^8 is 1/9 (x^2+2)^9+C