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

The integral of x/((x^2+1)(x^2+4)) is 1/6 ln|x^2+1|-1/6 ln|x^2+4|+C

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

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

\frac{1}{6} ln x^{2} + 1 - \frac{1}{6} ln x^{2} + 4 + C

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= \frac{1}{6} ln x^{2} + 1 - \frac{1}{6} ln x^{2} + 4

Add a constant to the solution

= \frac{1}{6} ln x^{2} + 1 - \frac{1}{6} ln x^{2} + 4 + C

\int (1 + 0.5 t) e^{- 0.004 t} d t

\frac{\partial}{\partial x} (5 x^{2} - 2 y)

\int (x^{\frac{3}{2}} + 8 x + 9) d x

f (x) = - 3 x^{2} + 6 x + 4, at x = 1

x \to 0 lim (\frac{ln ( x + 2 )}{x})

What is the integral of x/((x^2+1)(x^2+4)) ?
The integral of x/((x^2+1)(x^2+4)) is 1/6 ln|x^2+1|-1/6 ln|x^2+4|+C