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

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

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

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

\frac{1}{8} (x^{2} + 7)^{8} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{7}}{2} d u

= 2 \cdot \int \frac{u ^{7}}{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^{7} d u

Apply the Power Rule

: \frac{u ^{8}}{8}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{8}}{8}

Substitute back

u = x^{2} + 7

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{8}}{8}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 7 ) ^{8}}{8} : \frac{1}{8} (x^{2} + 7)^{8}

= \frac{1}{8} (x^{2} + 7)^{8}

Add a constant to the solution

= \frac{1}{8} (x^{2} + 7)^{8} + C

\frac{d y}{d x} = - \frac{1}{3} x^{3}

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

\int \frac{cos ( 2 x )}{cos ( x )} d x

8 x^{2} y^{^{'}} = y^{^{'}} + 8 x e^{- y}

f (x) = - 7 x - \frac{2}{x ^{7}}

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