What is the integral of 2x(1+2x)^4 ?

The integral of 2x(1+2x)^4 is x^2+(16x^3)/3+12x^4+(64x^5)/5+(16x^6)/3+C

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integral of 2x(1+2x)^4

\int 2 x (1 + 2 x)^{4} d x

x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3} + C

Solution steps

\int 2 x (1 + 2 x)^{4} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int x (1 + 2 x)^{4} d x

Expand

x (1 + 2 x)^{4} : x + 8 x^{2} + 24 x^{3} + 32 x^{4} + 16 x^{5}

= 2 \cdot \int x + 8 x^{2} + 24 x^{3} + 32 x^{4} + 16 x^{5} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int x d x + \int 8 x^{2} d x + \int 24 x^{3} d x + \int 32 x^{4} d x + \int 16 x^{5} d x)

\int x d x = \frac{x ^{2}}{2}

\int 8 x^{2} d x = \frac{8 x ^{3}}{3}

\int 24 x^{3} d x = 6 x^{4}

\int 32 x^{4} d x = \frac{32 x ^{5}}{5}

\int 16 x^{5} d x = \frac{8 x ^{6}}{3}

= 2 (\frac{x ^{2}}{2} + \frac{8 x ^{3}}{3} + 6 x^{4} + \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3})

Simplify

2 (\frac{x ^{2}}{2} + \frac{8 x ^{3}}{3} + 6 x^{4} + \frac{32 x ^{5}}{5} + \frac{8 x ^{6}}{3}) : x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3}

= x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3}

Add a constant to the solution

= x^{2} + \frac{16 x ^{3}}{3} + 12 x^{4} + \frac{64 x ^{5}}{5} + \frac{16 x ^{6}}{3} + C

1 - e^{- 2 t}

y^{^{'}} + \frac{y}{x} = 1

\frac{d y}{d x} = 2 x y + 5 x

\frac{d}{d x} (7^{x} ln (x))

\int \frac{y - 1}{y ^{2}} d y

What is the integral of 2x(1+2x)^4 ?
The integral of 2x(1+2x)^4 is x^2+(16x^3)/3+12x^4+(64x^5)/5+(16x^6)/3+C