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

The integral of (x+1)*e^{3x+1} is e(1/3 e^{3x}x+2/9 e^{3x})+C

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

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

e (\frac{1}{3} e^{3 x} x + \frac{2}{9} e^{3 x}) + C

Solution steps

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

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

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

Simplify

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

Apply Integration By Parts

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

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

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

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

Expand

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

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

Add a constant to the solution

= e (\frac{1}{3} e^{3 x} x + \frac{2}{9} e^{3 x}) + C

\int (\frac{x}{2 - x})^{2} d x

\int \frac{1}{( x ^{2} + 16 )} d x

\int_{0}^{x} t^{4} + t^{6} d t

y^{^{''}} - 2 y^{^{'}} - 2 y = 0

x \to n - lim ([x])

What is the integral of (x+1)*e^{3x+1} ?
The integral of (x+1)*e^{3x+1} is e(1/3 e^{3x}x+2/9 e^{3x})+C