What is the integral of 1+e^{-x}sin(4x) ?

The integral of 1+e^{-x}sin(4x) is x-(4e^{-x}cos(4x))/(17)-(e^{-x}sin(4x))/(17)+C

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integral of 1+e^{-x}sin(4x)

integral

1 + e^{- x} sin (4 x)

x - \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17} + C

Solution steps

\int 1 + e^{- x} sin (4 x) 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 1 d x + \int e^{- x} sin (4 x) d x

\int 1 d x = x

\int e^{- x} sin (4 x) d x = - \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17}

= x - \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17}

Add a constant to the solution

= x - \frac{4 e ^{- x} cos ( 4 x )}{17} - \frac{e ^{- x} sin ( 4 x )}{17} + C

2 x^{- 3}

(4 - x^{2})^{1.5}

(1 - x^{2})^{3}

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

\frac{e ^{\frac{1}{y}}}{y}

What is the integral of 1+e^{-x}sin(4x) ?
The integral of 1+e^{-x}sin(4x) is x-(4e^{-x}cos(4x))/(17)-(e^{-x}sin(4x))/(17)+C