What is the integral of (4x+3)e^{3x+1} ?

The integral of (4x+3)e^{3x+1} is e(4/3 e^{3x}x+5/9 e^{3x})+C

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integral of (4x+3)e^{3x+1}

\int (4 x + 3) e^{3 x + 1} d x

e (\frac{4}{3} e^{3 x} x + \frac{5}{9} e^{3 x}) + C

Solution steps

\int (4 x + 3) e^{3 x + 1} d x

(4 x + 3) e^{3 x + 1} = e^{1} (4 x + 3) e^{3 x}

= \int e^{1} (4 x + 3) e^{3 x} d x

Take the constant out:

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

= e^{1} \cdot \int (4 x + 3) e^{3 x} d x

Simplify

= e \cdot \int (4 x + 3) e^{3 x} d x

Apply Integration By Parts

: \frac{4}{3} e^{3 x} x + e^{3 x} - \int \frac{4}{3} e^{3 x} d x

= e (\frac{4}{3} e^{3 x} x + e^{3 x} - \int \frac{4}{3} e^{3 x} d x)

\int \frac{4}{3} e^{3 x} d x = \frac{4}{9} e^{3 x}

= e (\frac{4}{3} e^{3 x} x + e^{3 x} - \frac{4}{9} e^{3 x})

Add similar elements:

e^{3 x} - \frac{4}{9} e^{3 x} = \frac{5}{9} e^{3 x}

= e (\frac{4}{3} e^{3 x} x + \frac{5}{9} e^{3 x})

Add a constant to the solution

= e (\frac{4}{3} e^{3 x} x + \frac{5}{9} e^{3 x}) + C

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

y = - x^{2} + 12 x - 32, y = 0

x \to 4 lim (\frac{15}{x - 4})

(\frac{d y}{d x}) - y = 4 e^{x} + 24 e^{(4 x)}

\frac{d}{d x} (- \frac{1}{2} cot^{3} (x))

What is the integral of (4x+3)e^{3x+1} ?
The integral of (4x+3)e^{3x+1} is e(4/3 e^{3x}x+5/9 e^{3x})+C