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

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

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

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

\frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x + \frac{3}{4} e^{2 x} + C

Solution steps

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

Apply Integration By Parts

: \frac{1}{2} e^{2 x} (x^{2} + 1) - \int e^{2 x} x d x

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

\int e^{2 x} x d x = \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

= \frac{1}{2} e^{2 x} (x^{2} + 1) - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

Simplify

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

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x + \frac{3}{4} e^{2 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{2} e^{2 x} x + \frac{3}{4} e^{2 x} + C

f (x) = 5 e^{x}

\frac{d}{d x} (\frac{3}{4} x^{\frac{3}{2}})

\frac{1}{( 3 x ) ^{2}}

x \to 2 lim (3 x^{2} + 7 x - 1)

tan (t) - e^{- t}

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