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

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

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

integral

2 x^{3} (e^{x})^{- 2}

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

Solution steps

\int 2 x^{3} (e^{x})^{- 2} d x

Take the constant out:

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

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

Simplify

x^{3} (e^{x})^{- 2} : \frac{x ^{3}}{e ^{2 x}}

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{16} d u

= 2 \cdot \int \frac{e ^{u} u ^{3}}{16} d u

Take the constant out:

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

= 2 \cdot \frac{1}{16} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= 2 \cdot \frac{1}{16} (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= 2 \cdot \frac{1}{16} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - 2 x

= 2 \cdot \frac{1}{16} ((- 2 x)^{3} e^{- 2 x} - 3 ((- 2 x)^{2} e^{- 2 x} - 2 (e^{- 2 x} (- 2 x) - e^{- 2 x})))

Simplify

2 \cdot \frac{1}{16} ((- 2 x)^{3} e^{- 2 x} - 3 ((- 2 x)^{2} e^{- 2 x} - 2 (e^{- 2 x} (- 2 x) - e^{- 2 x}))) : - e^{- 2 x} x^{3} - \frac{3}{4} e^{- 2 x} (2 x^{2} + 2 x + 1)

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

Add a constant to the solution

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

x^{3} (a^{3} - x^{3})^{5}

x (1 + x)^{3}

\frac{( 5 x - 6 )}{( x ^{2} - 2 x + 2 )}

\frac{x + 1}{( x )}

\frac{e}{( ( e ) ^{x} ) ^{2}}

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