What is the integral of 2x(x^2+4)^6 ?

The integral of 2x(x^2+4)^6 is 1/7 (x^2+4)^7+C

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integral of 2x(x^2+4)^6

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

\frac{1}{7} (x^{2} + 4)^{7} + C

Solution steps

\int 2 x (x^{2} + 4)^{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} + 4)^{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} + 4

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{7}}{7}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{7}}{7} : \frac{1}{7} (x^{2} + 4)^{7}

= \frac{1}{7} (x^{2} + 4)^{7}

Add a constant to the solution

= \frac{1}{7} (x^{2} + 4)^{7} + C

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g (x) = \frac{1}{5 x ^{2} + 6 x}

\frac{\partial}{\partial x} (\frac{1}{y} cos (\frac{x}{y}))

θ \to 0 lim (\frac{1 - cos ( θ )}{sin ( θ )})

What is the integral of 2x(x^2+4)^6 ?
The integral of 2x(x^2+4)^6 is 1/7 (x^2+4)^7+C