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

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

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integral of 4/(x^3+x)

\int \frac{4}{x ^{3} + x} d x

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

Solution steps

\int \frac{4}{x ^{3} + x} d x

Take the constant out:

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

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

Take the partial fraction of

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

= 4 \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

= 4 (\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

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

Add a constant to the solution

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

f (x) = ln (x + 1 + x^{2})

x^{3} e^{- 3 x}

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

f (x) = (- x^{2} + 4)^{2}

y = \frac{2}{x - 2} (4.1)

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