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

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

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integral of x/((x^2+1)^2)+e^{-4x}

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

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

Solution steps

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

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

\int e^{- 4 x} d x = - \frac{1}{4} e^{- 4 x}

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

Add a constant to the solution

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

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