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

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

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

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

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

Solution steps

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

Take the partial fraction of

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

= 4 \cdot \int \frac{1}{4 x} - \frac{x}{4 ( 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

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

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{x}{4 ( x ^{2} + 4 )} d x = \frac{1}{8} ln x^{2} + 4

= 4 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

4 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

= ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 4

Add a constant to the solution

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

\int 0.2 e^{- 0.3 x} d x

\int \frac{1}{n + n} d n

\int - ln (x) d x

\int \frac{1}{( x - 1 ) ^{5}} d x

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

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