What is the integral of (8x^2+8x+2)/((4x^2+1)^2) ?

The integral of (8x^2+8x+2)/((4x^2+1)^2) is arctan(2x)-1/(4x^2+1)+C

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integral of (8x^2+8x+2)/((4x^2+1)^2)

\int \frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} d x

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

Solution steps

\int \frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} d x

Take the partial fraction of

\frac{8 x ^{2} + 8 x + 2}{( 4 x ^{2} + 1 ) ^{2}} : \frac{2}{4 x ^{2} + 1} + \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}}

= \int \frac{2}{4 x ^{2} + 1} + \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}} 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{2}{4 x ^{2} + 1} d x + \int \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}} d x

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

\int \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}} d x = - \frac{1}{4 x ^{2} + 1}

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

Add a constant to the solution

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

\int_{1}^{2} \frac{1}{x ^{3} + x} d x

f (x) = x^{2} - x + 2, at x = 2

\int_{0}^{1} \frac{e ^{x}}{1 + e ^{2 x}} d x

x \to 3 lim (\frac{1}{3} x^{2} - 4 x + 10)

\frac{\partial}{\partial y} (\frac{3 y}{x})

What is the integral of (8x^2+8x+2)/((4x^2+1)^2) ?
The integral of (8x^2+8x+2)/((4x^2+1)^2) is arctan(2x)-1/(4x^2+1)+C