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

The integral of 3/(x(x^2+1)) is 3(ln|x|-1/2 ln|x^2+1|)+C

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

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

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

Solution steps

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

Take the partial fraction of

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

= 3 \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

= 3 (\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

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

Add a constant to the solution

= 3 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

\frac{d}{d x} (∣ (- 3)^{x} ∣)

y^{^{'}} x^{2} + y x^{2} = x y^{5}

f (x) = \frac{1}{2}

\frac{2 x ^{2} - 1}{x + 1}

t^{3}

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