What is the integral of 9/(x(x^2+1)) ?

The integral of 9/(x(x^2+1)) is 9(ln|x|-1/2 ln|x^2+1|)+C

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integral of 9/(x(x^2+1))

\int \frac{9}{x ( x ^{2} + 1 )} d x

9 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

Solution steps

\int \frac{9}{x ( x ^{2} + 1 )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ( x ^{2} + 1 )} d x

Take the partial fraction of

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

= 9 \cdot \int \frac{1}{x} - \frac{x}{x ^{2} + 1} 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

= 9 (\int \frac{1}{x} d x - \int \frac{x}{x ^{2} + 1} d x)

\int \frac{1}{x} d x = ln ∣ x ∣

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

= 9 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

Add a constant to the solution

= 9 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

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

sec (\frac{5}{6}) x^{3}

\frac{\partial}{\partial x} (- e^{k z} A w sin (k x - wt))

\frac{\partial}{\partial x} (e^{4 x e^{y}})

x \to 1 - lim (\frac{2}{x} - \frac{1}{ln ( x )})

What is the integral of 9/(x(x^2+1)) ?
The integral of 9/(x(x^2+1)) is 9(ln|x|-1/2 ln|x^2+1|)+C