What is the integral of 1/(x^2+4)+cos(2x) ?

The integral of 1/(x^2+4)+cos(2x) is 1/2 arctan(x/2)+1/2 sin(2x)+C

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integral of 1/(x^2+4)+cos(2x)

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

\frac{1}{2} arctan (\frac{x}{2}) + \frac{1}{2} sin (2 x) + C

Solution steps

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

\int \frac{1}{x ^{2} + 4} d x = \frac{1}{2} arctan (\frac{x}{2})

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

= \frac{1}{2} arctan (\frac{x}{2}) + \frac{1}{2} sin (2 x)

Add a constant to the solution

= \frac{1}{2} arctan (\frac{x}{2}) + \frac{1}{2} sin (2 x) + C

\int \frac{x + 2}{x ^{2} - 3 x - 4} d x

\int x^{3} x - 1 d x

\int \frac{x ^{3} - 4 x ^{2} + 2 x}{x ^{2} + 1} d x

\int x^{2} + 2^{- x} d x

\int \frac{arctan ( 8 x )}{1 + 64 x ^{2}} d x

What is the integral of 1/(x^2+4)+cos(2x) ?
The integral of 1/(x^2+4)+cos(2x) is 1/2 arctan(x/2)+1/2 sin(2x)+C