What is the integral of 2x(x^2+4)^5 ?

The integral of 2x(x^2+4)^5 is 1/6 (x^2+4)^6+C

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integral of 2x(x^2+4)^5

\int 2 x (x^{2} + 4)^{5} d x

\frac{1}{6} (x^{2} + 4)^{6} + C

Solution steps

\int 2 x (x^{2} + 4)^{5} 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)^{5} d x

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= 2 \cdot \int \frac{u ^{5}}{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^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 2 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} + 4

= 2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{6}}{6}

Simplify

2 \cdot \frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{6}}{6} : \frac{1}{6} (x^{2} + 4)^{6}

= \frac{1}{6} (x^{2} + 4)^{6}

Add a constant to the solution

= \frac{1}{6} (x^{2} + 4)^{6} + C

y = \frac{6 x}{e ^{x}}

\frac{d}{d x} ((x - 1)^{\frac{3}{2}})

\frac{d}{d x} (\frac{x ^{3} + 2}{cos ( x )})

\frac{d}{d x} (arcsin (x^{3} y))

\int e^{2 x} d x

What is the integral of 2x(x^2+4)^5 ?
The integral of 2x(x^2+4)^5 is 1/6 (x^2+4)^6+C