What is the integral of 2te^t ?

The integral of 2te^t is 2(e^tt-e^t)+C

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integral of 2te^t

\int 2 t e^{t} d t

2 (e^{t} t - e^{t}) + C

Solution steps

\int 2 t e^{t} d t

Take the constant out:

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

= 2 \cdot \int t e^{t} d t

Apply Integration By Parts

: e^{t} t - \int e^{t} d t

= 2 (e^{t} t - \int e^{t} d t)

\int e^{t} d t = e^{t}

= 2 (e^{t} t - e^{t})

Add a constant to the solution

= 2 (e^{t} t - e^{t}) + C

\int (A + BT + C T^{2} + D T^{2}) \frac{1}{T} d T

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

\int \frac{x ^{5} + 4}{x ^{3} + x ^{2}} d x

\int_{0}^{R} R^{2} - x^{2} d x

y^{^{'}} - 3 y = 0