What is the integral of 3/(x(x^4+1)) ?

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

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integral of 3/(x(x^4+1))

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 1 + x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial z} (cos (x y))

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

e^{x} \cdot ln (x)

\frac{1}{1 + x ^{2}}

\int 17 e^{- 17 x} d x

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