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

The integral of e^{3x}2x is 2/9 (3e^{3x}x-e^{3x})+C

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integral of e^{3x}2x

\int e^{3 x} 2 x d x

\frac{2}{9} (3 e^{3 x} x - e^{3 x}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \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

= 2 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

= 2 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 2 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 x

= 2 \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Simplify

2 \cdot \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) : \frac{2}{9} (3 e^{3 x} x - e^{3 x})

= \frac{2}{9} (3 e^{3 x} x - e^{3 x})

Add a constant to the solution

= \frac{2}{9} (3 e^{3 x} x - e^{3 x}) + C

\int \frac{1}{( x ^{2} + 1 ) ^{1.5}} d x

\int 9 x x^{2} + 8 d x

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

\int (x^{2}) (4 - x^{3})^{4} d x

\int \frac{x ^{3} + 2 x ^{2}}{x ^{2}} d x

What is the integral of e^{3x}2x ?
The integral of e^{3x}2x is 2/9 (3e^{3x}x-e^{3x})+C