What is the integral of e^{3x}(2x^2+1) ?

The integral of e^{3x}(2x^2+1) is 2/3 e^{3x}x^2-4/9 e^{3x}x+13/27 e^{3x}+C

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integral of e^{3x}(2x^2+1)

\int e^{3 x} (2 x^{2} + 1) d x

\frac{2}{3} e^{3 x} x^{2} - \frac{4}{9} e^{3 x} x + \frac{13}{27} e^{3 x} + C

Solution steps

\int e^{3 x} (2 x^{2} + 1) d x

Expand

e^{3 x} (2 x^{2} + 1) : 2 e^{3 x} x^{2} + e^{3 x}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 2 e^{3 x} x^{2} d x = 2 (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x}))

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

= 2 (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x})) + \frac{1}{3} e^{3 x}

Simplify

2 (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x})) + \frac{1}{3} e^{3 x} : \frac{2}{3} e^{3 x} x^{2} - \frac{4}{9} e^{3 x} x + \frac{13}{27} e^{3 x}

= \frac{2}{3} e^{3 x} x^{2} - \frac{4}{9} e^{3 x} x + \frac{13}{27} e^{3 x}

Add a constant to the solution

= \frac{2}{3} e^{3 x} x^{2} - \frac{4}{9} e^{3 x} x + \frac{13}{27} e^{3 x} + C

What is the integral of e^{3x}(2x^2+1) ?
The integral of e^{3x}(2x^2+1) is 2/3 e^{3x}x^2-4/9 e^{3x}x+13/27 e^{3x}+C