What is the integral of e^{-1}f(x) ?

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

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integral of e^{-1}f(x)

integral

e^{- 1} f (x)

\frac{x}{e} \cdot \frac{f ^{2}}{2} + C

Solution steps

\int e^{- 1} f x df

Take the constant out:

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

= x e^{- 1} \cdot \int fdf

Simplify

x e^{- 1} : \frac{x}{e}

= \frac{x}{e} \cdot \int fdf

Apply the Power Rule

: \frac{f ^{2}}{2}

= \frac{x}{e} \cdot \frac{f ^{2}}{2}

Add a constant to the solution

= \frac{x}{e} \cdot \frac{f ^{2}}{2} + C

tan^{5} (x)

x^{2} (1 + x^{3})

\frac{1}{( x + 1 ) ( x ^{2} + 1 ) ^{2}}

d^{2} (v) v

\frac{1}{( 2 + x )}

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