What is the integral of 7/(x(x^4+6)) ?

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

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integral of 7/(x(x^4+6))

\int \frac{7}{x ( x ^{4} + 6 )} d x

\frac{7}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6) + C

Solution steps

\int \frac{7}{x ( x ^{4} + 6 )} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{x ( x ^{4} + 6 )} d x

Apply u-substitution

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

= 7 \cdot \int \frac{1}{4 u ( u + 6 )} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 6 )} d u

Take the partial fraction of

\frac{1}{u ( u + 6 )} : \frac{1}{6 u} - \frac{1}{6 ( u + 6 )}

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{6 u} - \frac{1}{6 ( u + 6 )} 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

= 7 \cdot \frac{1}{4} (\int \frac{1}{6 u} d u - \int \frac{1}{6 ( u + 6 )} d u)

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

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

= 7 \cdot \frac{1}{4} (\frac{1}{6} ln ∣ u ∣ - \frac{1}{6} ln ∣ u + 6 ∣)

Substitute back

u = x^{4}

= 7 \cdot \frac{1}{4} (\frac{1}{6} ln x^{4} - \frac{1}{6} ln x^{4} + 6)

Simplify

7 \cdot \frac{1}{4} (\frac{1}{6} ln x^{4} - \frac{1}{6} ln x^{4} + 6) : \frac{7}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6)

= \frac{7}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6)

Add a constant to the solution

= \frac{7}{4} (\frac{2}{3} ln (x) - \frac{1}{6} ln x^{4} + 6) + C

k = 1 \sum \infty \frac{1}{2 ^{k}}

3 x + y - 2 + \frac{d y}{d x} (x - 1) = 0

n = 0 \sum \infty \frac{1}{2 n + 1}

\int \frac{1}{1 + 4 x ^{2} + 4 x ^{4}} d x

x \to \frac{π}{4} lim (2 sin (3 x))

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