What is the integral of 5/(x(x^4+7)) ?

The integral of 5/(x(x^4+7)) is 5/7 ln(x)-5/28 ln|x^4+7|+C

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integral of 5/(x(x^4+7))

\int \frac{5}{x ( x ^{4} + 7 )} d x

\frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7 + C

Solution steps

\int \frac{5}{x ( x ^{4} + 7 )} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x ( x ^{4} + 7 )} d x

Apply u-substitution

: \int \frac{1}{4 u ( u + 7 )} d u

= 5 \cdot \int \frac{1}{4 u ( u + 7 )} d u

Take the constant out:

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

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{u ( u + 7 )} d u

Take the partial fraction of

\frac{1}{u ( u + 7 )} : \frac{1}{7 u} - \frac{1}{7 ( u + 7 )}

= 5 \cdot \frac{1}{4} \cdot \int \frac{1}{7 u} - \frac{1}{7 ( u + 7 )} 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

= 5 \cdot \frac{1}{4} (\int \frac{1}{7 u} d u - \int \frac{1}{7 ( u + 7 )} d u)

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{1}{7 ( u + 7 )} d u = \frac{1}{7} ln ∣ u + 7 ∣

= 5 \cdot \frac{1}{4} (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = x^{4}

= 5 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7)

Simplify

5 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7) : \frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7

= \frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7

Add a constant to the solution

= \frac{5}{7} ln (x) - \frac{5}{28} ln x^{4} + 7 + C

\frac{d}{d x} (\frac{2 + e ^{x}}{2 - e ^{- x}})

\int \frac{x - 2}{x + 1} d x

h \to 0 lim (\frac{f ( h ) ( 8 + h ) - f ( 8 )}{h})

(- 5, - 10), (- 1, 5)

\frac{( 2 - 5 s )}{s ^{2} + 9}

What is the integral of 5/(x(x^4+7)) ?
The integral of 5/(x(x^4+7)) is 5/7 ln(x)-5/28 ln|x^4+7|+C