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

The integral of 3/(x(x^4+7)) is 3/7 ln(x)-3/28 ln|x^4+7|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{3}{7} ln (x) - \frac{3}{28} ln x^{4} + 7 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \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

= 3 \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 )}

= 3 \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

= 3 \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 ∣

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

Substitute back

u = x^{4}

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

Simplify

3 \cdot \frac{1}{4} (\frac{1}{7} ln x^{4} - \frac{1}{7} ln x^{4} + 7) : \frac{3}{7} ln (x) - \frac{3}{28} ln x^{4} + 7

= \frac{3}{7} ln (x) - \frac{3}{28} ln x^{4} + 7

Add a constant to the solution

= \frac{3}{7} ln (x) - \frac{3}{28} ln x^{4} + 7 + C

\int e^{7 x + 8} d x

\int (e^{x} sin (y) - y) d x

ln (tan^{2} (x))

y = x, y = x^{2} - 2, y = 0

x y^{^{'}} - 2 y = x^{4}

What is the integral of 3/(x(x^4+7)) ?
The integral of 3/(x(x^4+7)) is 3/7 ln(x)-3/28 ln|x^4+7|+C