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

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

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

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

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

Solution steps

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

Apply u-substitution

= \int \frac{cot ( ln ( u ) )}{4 u} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cot ( ln ( u ) )}{u} d u

Apply u-substitution

= \frac{1}{4} \cdot \int cot (v) d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int \frac{cos ( v )}{sin ( v )} d v

Apply u-substitution

= \frac{1}{4} \cdot \int \frac{1}{w} d w

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= \frac{1}{4} ln ∣ w ∣

Substitute back

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

Add a constant to the solution

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

\int \frac{1}{x ^{2} + 12} d x

3 x, 9 x^{2}

n = 1 \sum \infty \frac{2 n}{1 1 ^{n}}

\frac{d}{d x} (\frac{4 2}{3} x^{\frac{3}{2}} - 1)

(1 - x^{\frac{2}{3}})^{\frac{3}{2}}

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