What is the integral of 2x(x^2+6)^6 ?

The integral of 2x(x^2+6)^6 is 1/7 (x^2+6)^7+C

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integral of 2x(x^2+6)^6

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

\frac{1}{7} (x^{2} + 6)^{7} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{6}}{2} d u

= 2 \cdot \int \frac{u ^{6}}{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^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{7}}{7}

Substitute back

u = x^{2} + 6

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{7}}{7}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 6 ) ^{7}}{7} : \frac{1}{7} (x^{2} + 6)^{7}

= \frac{1}{7} (x^{2} + 6)^{7}

Add a constant to the solution

= \frac{1}{7} (x^{2} + 6)^{7} + C

x \to - 0.2 lim (12.1951)

e^{e^{x}}

t^{2} y^{^{''}} + t y^{^{'}} + 81 y = 0

\frac{\partial}{\partial y} (x + sin (y))

x \to 0 lim (\frac{e ^{x} - 1}{x ^{7}})

What is the integral of 2x(x^2+6)^6 ?
The integral of 2x(x^2+6)^6 is 1/7 (x^2+6)^7+C