What is the integral of 9/(x(x^4+7)) ?

The integral of 9/(x(x^4+7)) is 9/7 ln(x)-9/28 ln|x^4+7|+C

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integral of 9/(x(x^4+7))

\int \frac{9}{x ( x ^{4} + 7 )} d x

\frac{9}{7} ln (x) - \frac{9}{28} ln x^{4} + 7 + C

Solution steps

\int \frac{9}{x ( x ^{4} + 7 )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x ( x ^{4} + 7 )} d x

Apply u-substitution

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

= 9 \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

= 9 \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 )}

= 9 \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

= 9 \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 ∣

= 9 \cdot \frac{1}{4} (\frac{1}{7} ln ∣ u ∣ - \frac{1}{7} ln ∣ u + 7 ∣)

Substitute back

u = x^{4}

= 9 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7)

Simplify

9 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7) : \frac{9}{7} ln (x) - \frac{9}{28} ln x^{4} + 7

= \frac{9}{7} ln (x) - \frac{9}{28} ln x^{4} + 7

Add a constant to the solution

= \frac{9}{7} ln (x) - \frac{9}{28} ln x^{4} + 7 + C

\int 2 x^{2} cos (x^{3}) d x

f (x) = \frac{1}{x ^{3} + 2 x + 1}

ln (x^{3} + 3 x^{2} + 5 x + 6)

\frac{d}{d x} (2 x y - 1)

x \to 3 lim (2 x - 8)

What is the integral of 9/(x(x^4+7)) ?
The integral of 9/(x(x^4+7)) is 9/7 ln(x)-9/28 ln|x^4+7|+C