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

The integral of ((x^2+1))/(e^x) is -(x^2+1)/(e^x)+2(-x/(e^x)-1/(e^x))+C

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integral of ((x^2+1))/(e^x)

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

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

Solution steps

\int \frac{x ^{2} + 1}{e ^{x}} d x

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

y = 0, y = x, [0, 1]

f (x) = \frac{5}{x + 4}

\int_{0}^{t} t^{2} e^{- \frac{t}{4}} d t

\int (x + 6) (3 x - 4) d x

\int_{- 2}^{0} 3 x 4 - x^{2} d x

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