What is the integral of 18xe^{3x+1} ?

The integral of 18xe^{3x+1} is 18(1/3 e^{3x+1}x-1/9 e^{3x+1})+C

integral of 18xe^{3x+1}

\int 18 x e^{3 x + 1} d x

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

Solution steps

\int 18 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

= 18 \cdot \int x e^{3 x + 1} d x

Apply Integration By Parts

= 18 (\frac{1}{3} e^{3 x + 1} x - \int \frac{1}{3} e^{3 x + 1} d x)

\int \frac{1}{3} e^{3 x + 1} d x = \frac{1}{9} e^{3 x + 1}

= 18 (\frac{1}{3} e^{3 x + 1} x - \frac{1}{9} e^{3 x + 1})

Add a constant to the solution

= 18 (\frac{1}{3} e^{3 x + 1} x - \frac{1}{9} e^{3 x + 1}) + C

\int \frac{4 x ^{2} + 6}{x ( x ^{2} + 3 )} d x

\int \frac{4 x ^{2} + 3 x + 19}{( x + 1 ) ( x ^{2} + 4 )} d x

\int \frac{39 x ^{2}}{( x + 1 ) ( x ^{2} + 1 )} d x

\int sin^{2} (x) \cdot x d x

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

What is the integral of 18xe^{3x+1} ?
The integral of 18xe^{3x+1} is 18(1/3 e^{3x+1}x-1/9 e^{3x+1})+C