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

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

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integral of e^{-x}sin(x)+1

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

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

Solution steps

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d x = x

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

Add a constant to the solution

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

\int \frac{1}{( x + 2 ) ( x - 5 )} d x

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

n = 1 \sum \infty \frac{2 n}{3 n + 1}

2 x^{2} \frac{d y}{d x} = 3 x y + y^{2}

\frac{d}{d t} (cos (t) - t sin (t))

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