What is the integral of (t-x)^3e^x ?

The integral of (t-x)^3e^x is (e^x(-x+t)^4)/4+C

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integral of (t-x)^3e^x

integral

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

\frac{e ^{x} ( - x + t ) ^{4}}{4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u^{3} d u

= e^{x} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= e^{x} \frac{u ^{4}}{4}

Substitute back

u = - x + t

= e^{x} \frac{( - x + t ) ^{4}}{4}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{e ^{x} ( - x + t ) ^{4}}{4}

Add a constant to the solution

= \frac{e ^{x} ( - x + t ) ^{4}}{4} + C

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

\frac{1}{( y ^{2} - y )}

\frac{1}{a - x}

(e^{x})^{- 1}

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

What is the integral of (t-x)^3e^x ?
The integral of (t-x)^3e^x is (e^x(-x+t)^4)/4+C