What is the integral of 3/((x^2+1)(x^2+4)) ?

The integral of 3/((x^2+1)(x^2+4)) is arctan(x)-1/2 arctan(x/2)+C

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integral of 3/((x^2+1)(x^2+4))

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} d x

Take the partial fraction of

\frac{1}{( x ^{2} + 1 ) ( x ^{2} + 4 )} : \frac{1}{3 ( x ^{2} + 1 )} - \frac{1}{3 ( x ^{2} + 4 )}

= 3 \cdot \int \frac{1}{3 ( x ^{2} + 1 )} - \frac{1}{3 ( x ^{2} + 4 )} 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

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

\int \frac{1}{3 ( x ^{2} + 1 )} d x = \frac{1}{3} arctan (x)

\int \frac{1}{3 ( x ^{2} + 4 )} d x = \frac{1}{6} arctan (\frac{x}{2})

= 3 (\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2}))

Simplify

3 (\frac{1}{3} arctan (x) - \frac{1}{6} arctan (\frac{x}{2})) : arctan (x) - \frac{1}{2} arctan (\frac{x}{2})

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

Add a constant to the solution

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

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What is the integral of 3/((x^2+1)(x^2+4)) ?
The integral of 3/((x^2+1)(x^2+4)) is arctan(x)-1/2 arctan(x/2)+C