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

The integral of (2+x/(2x^2+1)) 1/(2x^2+1) is sqrt(2)arctan(sqrt(2)x)-1/(4(2x^2+1))+C

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

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

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

Solution steps

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

Simplify

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

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

Expand

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

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

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

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

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

Add a constant to the solution

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

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