What is the integral of 7xe^{(3x-1)} ?

The integral of 7xe^{(3x-1)} is 7/9 e^{3x-1}(3x-1)+C

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integral of 7xe^{(3x-1)}

\int 7 x e^{(3 x - 1)} d x

\frac{7}{9} e^{3 x - 1} (3 x - 1) + C

Solution steps

\int 7 x e^{3 x - 1} d x

Take the constant out:

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

= 7 \cdot \int x e^{3 x - 1} d x

x e^{3 x - 1} = e^{- 1} x e^{3 x}

= 7 \cdot \int e^{- 1} x e^{3 x} d x

Take the constant out:

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

= 7 e^{- 1} \cdot \int x e^{3 x} d x

Apply exponent rule:

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

= 7 \cdot \frac{1}{e} \cdot \int x e^{3 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= 7 \cdot \frac{1}{e} \cdot \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= 7 \cdot \frac{1}{e} \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 7 \cdot \frac{1}{e} \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 7 \cdot \frac{1}{e} \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 x

= 7 \cdot \frac{1}{e} \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Simplify

7 \cdot \frac{1}{e} \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) : \frac{7}{9} e^{3 x - 1} (3 x - 1)

= \frac{7}{9} e^{3 x - 1} (3 x - 1)

Add a constant to the solution

= \frac{7}{9} e^{3 x - 1} (3 x - 1) + C

n = 1 \sum \infty \frac{n !}{10 8 ^{n}}

y + x y^{^{'}} = a (1 + x y)

\frac{d}{d x} ((2 x + 5)^{7})

y^{^{'}} = \frac{12 ( 2 y - 1 )}{( 3 x - 2 ) ^{2}}, y (2) = 0

\int x^{3} cos (3 x) d x

What is the integral of 7xe^{(3x-1)} ?
The integral of 7xe^{(3x-1)} is 7/9 e^{3x-1}(3x-1)+C