What is the integral of (-1)/4 e^{-4x}3x^2 ?

The integral of (-1)/4 e^{-4x}3x^2 is 3/128 e^{-4x}(8x^2+4x+1)+C

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integral of (-1)/4 e^{-4x}3x^2

\int \frac{- 1}{4} e^{- 4 x} 3 x^{2} d x

\frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1) + C

Solution steps

\int \frac{- 1}{4} e^{- 4 x} \cdot 3 x^{2} d x

Take the constant out:

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

= \frac{- 1}{4} \cdot 3 \cdot \int e^{- 4 x} x^{2} d x

\frac{- 1}{4} \cdot 3 = - \frac{3}{4}

= (- \frac{3}{4}) \cdot \int e^{- 4 x} x^{2} d x

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{64} d u

= (- \frac{3}{4}) \cdot \int - \frac{e ^{u} u ^{2}}{64} d u

Take the constant out:

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

= (- \frac{3}{4}) (- \frac{1}{64} \cdot \int e^{u} u^{2} d u)

Apply Integration By Parts

: u^{2} e^{u} - \int 2 u e^{u} d u

= (- \frac{3}{4}) (- \frac{1}{64} (u^{2} e^{u} - \int 2 u e^{u} d u))

Take the constant out:

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

= (- \frac{3}{4}) (- \frac{1}{64} (u^{2} e^{u} - 2 \cdot \int u e^{u} d u))

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

= (- \frac{3}{4}) (- \frac{1}{64} (u^{2} e^{u} - 2 (e^{u} u - \int e^{u} d u)))

Use the common integral:

\int e^{u} d u = e^{u}

= (- \frac{3}{4}) (- \frac{1}{64} (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - 4 x

= (- \frac{3}{4}) (- \frac{1}{64} ((- 4 x)^{2} e^{- 4 x} - 2 (e^{- 4 x} (- 4 x) - e^{- 4 x})))

Simplify

(- \frac{3}{4}) (- \frac{1}{64} ((- 4 x)^{2} e^{- 4 x} - 2 (e^{- 4 x} (- 4 x) - e^{- 4 x}))) : \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1)

= \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1)

Add a constant to the solution

= \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1) + C

\int 15 (x + 2)^{4} d x

y^{^{'}} = y (1 - y), y (0) = - 5

n = 0 \sum \infty \frac{1}{( 1 + n ) ^{2}}

\frac{d}{d x} (3 cos (3 x))

4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 5 x

What is the integral of (-1)/4 e^{-4x}3x^2 ?
The integral of (-1)/4 e^{-4x}3x^2 is 3/128 e^{-4x}(8x^2+4x+1)+C