What is the integral of 2e^{tx}x ?

The integral of 2e^{tx}x is 2(e^{tx} 1/t x-e^{tx} 1/(t^2))+C

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integral of 2e^{tx}x

\int 2 e^{t x} x d x

2 (e^{t x} \frac{1}{t} x - e^{t x} \frac{1}{t ^{2}}) + C

Solution steps

\int 2 e^{t x} x d x

Take the constant out:

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

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

Apply Integration By Parts

: e^{t x} \frac{1}{t} x - \int e^{t x} \frac{1}{t} d x

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

\int e^{t x} \frac{1}{t} d x = e^{t x} \frac{1}{t ^{2}}

= 2 (e^{t x} \frac{1}{t} x - e^{t x} \frac{1}{t ^{2}})

Add a constant to the solution

= 2 (e^{t x} \frac{1}{t} x - e^{t x} \frac{1}{t ^{2}}) + C

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What is the integral of 2e^{tx}x ?
The integral of 2e^{tx}x is 2(e^{tx} 1/t x-e^{tx} 1/(t^2))+C