What is the integral of 1/((arctan(y/4))(16+y^2)) ?

The integral of 1/((arctan(y/4))(16+y^2)) is 1/4 ln|arctan(y/4)|+C

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integral of 1/((arctan(y/4))(16+y^2))

\int \frac{1}{( arctan ( \frac{y}{4} ) ) ( 16 + y ^{2} )} d y

\frac{1}{4} ln arctan (\frac{y}{4}) + C

Solution steps

\int \frac{1}{arctan ( \frac{y}{4} ) ( 16 + y ^{2} )} d y

Apply Integral Substitution

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

Apply u-substitution

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{4} ln ∣ v ∣

Substitute back

= \frac{1}{4} ln arctan (\frac{y}{4})

Add a constant to the solution

= \frac{1}{4} ln arctan (\frac{y}{4}) + C

What is the integral of 1/((arctan(y/4))(16+y^2)) ?
The integral of 1/((arctan(y/4))(16+y^2)) is 1/4 ln|arctan(y/4)|+C