What is the integral of 7/(x(x^4+3)) ?

The integral of 7/(x(x^4+3)) is 7/3 ln(x)-7/12 ln|x^4+3|+C

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integral of 7/(x(x^4+3))

\int \frac{7}{x ( x ^{4} + 3 )} d x

\frac{7}{3} ln (x) - \frac{7}{12} ln x^{4} + 3 + C

Solution steps

\int \frac{7}{x ( x ^{4} + 3 )} 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} + 3 )} d x

Apply u-substitution

: \int \frac{1}{4 u ( u + 3 )} d u

= 7 \cdot \int \frac{1}{4 u ( u + 3 )} 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 + 3 )} d u

Take the partial fraction of

\frac{1}{u ( u + 3 )} : \frac{1}{3 u} - \frac{1}{3 ( u + 3 )}

= 7 \cdot \frac{1}{4} \cdot \int \frac{1}{3 u} - \frac{1}{3 ( u + 3 )} 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}{3 u} d u - \int \frac{1}{3 ( u + 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

\int \frac{1}{3 ( u + 3 )} d u = \frac{1}{3} ln ∣ u + 3 ∣

= 7 \cdot \frac{1}{4} (\frac{1}{3} ln ∣ u ∣ - \frac{1}{3} ln ∣ u + 3 ∣)

Substitute back

u = x^{4}

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

Simplify

7 \cdot \frac{1}{4} (\frac{1}{3} ln x^{4} - \frac{1}{3} ln x^{4} + 3) : \frac{7}{3} ln (x) - \frac{7}{12} ln x^{4} + 3

= \frac{7}{3} ln (x) - \frac{7}{12} ln x^{4} + 3

Add a constant to the solution

= \frac{7}{3} ln (x) - \frac{7}{12} ln x^{4} + 3 + C

\frac{5}{s ( s + 3 )}

\int (cos (x)) (sin (x)) d x

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 2}{x})

x^{2} \frac{d y}{d x} + sin (x) - y = 0

x \to + 0 lim (3 x^{2})

What is the integral of 7/(x(x^4+3)) ?
The integral of 7/(x(x^4+3)) is 7/3 ln(x)-7/12 ln|x^4+3|+C