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

The integral of (2t^2-1)e^{3t+1} is 2/3 e^{3t+1}t^2-4/9 e^{3t+1}t-5/27 e^{3t+1}+C

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

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

\frac{2}{3} e^{3 t + 1} t^{2} - \frac{4}{9} e^{3 t + 1} t - \frac{5}{27} e^{3 t + 1} + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{3} e^{3 t + 1} (2 t^{2} - 1) - \int \frac{4}{3} e^{3 t + 1} t d t

= \frac{1}{3} e^{3 t + 1} (2 t^{2} - 1) - \int \frac{4}{3} e^{3 t + 1} t d t

\int \frac{4}{3} e^{3 t + 1} t d t = \frac{4}{3} (\frac{1}{3} e^{3 t + 1} t - \frac{1}{9} e^{3 t + 1})

= \frac{1}{3} e^{3 t + 1} (2 t^{2} - 1) - \frac{4}{3} (\frac{1}{3} e^{3 t + 1} t - \frac{1}{9} e^{3 t + 1})

Simplify

\frac{1}{3} e^{3 t + 1} (2 t^{2} - 1) - \frac{4}{3} (\frac{1}{3} e^{3 t + 1} t - \frac{1}{9} e^{3 t + 1}) : \frac{2}{3} e^{3 t + 1} t^{2} - \frac{4}{9} e^{3 t + 1} t - \frac{5}{27} e^{3 t + 1}

= \frac{2}{3} e^{3 t + 1} t^{2} - \frac{4}{9} e^{3 t + 1} t - \frac{5}{27} e^{3 t + 1}

Add a constant to the solution

= \frac{2}{3} e^{3 t + 1} t^{2} - \frac{4}{9} e^{3 t + 1} t - \frac{5}{27} e^{3 t + 1} + C

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What is the integral of (2t^2-1)e^{3t+1} ?
The integral of (2t^2-1)e^{3t+1} is 2/3 e^{3t+1}t^2-4/9 e^{3t+1}t-5/27 e^{3t+1}+C