What is the integral of 2x(x^2+7)^8 ?

The integral of 2x(x^2+7)^8 is 1/9 (x^2+7)^9+C

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integral of 2x(x^2+7)^8

\int 2 x (x^{2} + 7)^{8} d x

\frac{1}{9} (x^{2} + 7)^{9} + C

Solution steps

\int 2 x (x^{2} + 7)^{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} + 7)^{8} d x

Apply u-substitution

= 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

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

Substitute back

u = x^{2} + 7

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{9}}{9}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{9}}{9} : \frac{1}{9} (x^{2} + 7)^{9}

= \frac{1}{9} (x^{2} + 7)^{9}

Add a constant to the solution

= \frac{1}{9} (x^{2} + 7)^{9} + C

What is the integral of 2x(x^2+7)^8 ?
The integral of 2x(x^2+7)^8 is 1/9 (x^2+7)^9+C