What is the integral of (1+4cot(x))/(4-cot(x)) ?

The integral of (1+4cot(x))/(4-cot(x)) is ln|-4+cot(x)|-1/2 ln(csc^2(x))+C

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integral of (1+4cot(x))/(4-cot(x))

\int \frac{1 + 4 cot ( x )}{4 - cot ( x )} d x

ln ∣ - 4 + cot (x) ∣ - \frac{1}{2} ln (csc^{2} (x)) + C

Solution steps

\int \frac{1 + 4 cot ( x )}{4 - cot ( x )} d x

Apply Trigonometric Substitution

: \int - \frac{1 + 4 u}{( 4 - u ) ( 1 + u ^{2} )} d u

= \int - \frac{1 + 4 u}{( 4 - u ) ( 1 + u ^{2} )} d u

Take the constant out:

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

= - \int \frac{1 + 4 u}{( 4 - u ) ( 1 + u ^{2} )} d u

Take the partial fraction of

\frac{1 + 4 u}{( 4 - u ) ( 1 + u ^{2} )} : - \frac{1}{u - 4} + \frac{u}{u ^{2} + 1}

= - \int - \frac{1}{u - 4} + \frac{u}{u ^{2} + 1} 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

= - (- \int \frac{1}{u - 4} d u + \int \frac{u}{u ^{2} + 1} d u)

\int \frac{1}{u - 4} d u = ln ∣ u - 4 ∣

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

= - (- ln ∣ u - 4 ∣ + \frac{1}{2} ln u^{2} + 1)

Substitute back

u = cot (x)

= - (- ln ∣ cot (x) - 4 ∣ + \frac{1}{2} ln cot^{2} (x) + 1)

Simplify

- (- ln ∣ cot (x) - 4 ∣ + \frac{1}{2} ln cot^{2} (x) + 1) : ln ∣ - 4 + cot (x) ∣ - \frac{1}{2} ln (csc^{2} (x))

= ln ∣ - 4 + cot (x) ∣ - \frac{1}{2} ln (csc^{2} (x))

Add a constant to the solution

= ln ∣ - 4 + cot (x) ∣ - \frac{1}{2} ln (csc^{2} (x)) + C

\int cot^{3} (d) x d x

\int (cot^{2} (x) + 1)^{2} csc^{2} (x) d x

\int x (2 x + 3)^{7} d x

\int a x^{7} d x

\int \frac{cot ( t ) csc ( t )}{csc ^{7} ( t )} d t

What is the integral of (1+4cot(x))/(4-cot(x)) ?
The integral of (1+4cot(x))/(4-cot(x)) is ln|-4+cot(x)|-1/2 ln(csc^2(x))+C