What is the integral of 1/(x^7+4x) ?

The integral of 1/(x^7+4x) is 1/6 (3/2 ln(x)-1/4 ln|x^6+4|)+C

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integral of 1/(x^7+4x)

\int \frac{1}{x ^{7} + 4 x} d x

\frac{1}{6} (\frac{3}{2} ln (x) - \frac{1}{4} ln x^{6} + 4) + C

Solution steps

\int \frac{1}{x ^{7} + 4 x} d x

Factor

x^{7} + 4 x : x (x^{6} + 4)

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

Apply u-substitution

= \int \frac{1}{6 u ( u + 4 )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u ( u + 4 )} d u

Take the partial fraction of

\frac{1}{u ( u + 4 )} : \frac{1}{4 u} - \frac{1}{4 ( u + 4 )}

= \frac{1}{6} \cdot \int \frac{1}{4 u} - \frac{1}{4 ( u + 4 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{6} (\int \frac{1}{4 u} d u - \int \frac{1}{4 ( u + 4 )} d u)

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

\int \frac{1}{4 ( u + 4 )} d u = \frac{1}{4} ln ∣ u + 4 ∣

= \frac{1}{6} (\frac{1}{4} ln ∣ u ∣ - \frac{1}{4} ln ∣ u + 4 ∣)

Substitute back

u = x^{6}

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

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

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

Add a constant to the solution

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

What is the integral of 1/(x^7+4x) ?
The integral of 1/(x^7+4x) is 1/6 (3/2 ln(x)-1/4 ln|x^6+4|)+C