What is the integral of 4/(5xsqrt(x^2+9)) ?

The integral of 4/(5xsqrt(x^2+9)) is 4/15 ln|tan(1/2 arctan(1/3 x))|+C

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integral of 4/(5xsqrt(x^2+9))

\int \frac{4}{5 x x ^{2} + 9} d x

\frac{4}{15} ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + C

Solution steps

\int \frac{4}{5 x x ^{2} + 9} d x

Take the constant out:

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

= \frac{4}{5} \cdot \int \frac{1}{x x ^{2} + 9} d x

Apply Trigonometric Substitution

= \frac{4}{5} \cdot \int \frac{sec ( u )}{3 tan ( u )} d u

Take the constant out:

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

= \frac{4}{5} \cdot \frac{1}{3} \cdot \int \frac{sec ( u )}{tan ( u )} d u

Apply u-substitution

= \frac{4}{5} \cdot \frac{1}{3} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{4}{5} \cdot \frac{1}{3} ln ∣ v ∣

Substitute back

= \frac{4}{5} \cdot \frac{1}{3} ln tan (\frac{arctan ( \frac{1}{3} x )}{2})

Simplify

\frac{4}{5} \cdot \frac{1}{3} ln tan (\frac{arctan ( \frac{1}{3} x )}{2}) : \frac{4}{15} ln tan (\frac{1}{2} arctan (\frac{1}{3} x))

= \frac{4}{15} ln tan (\frac{1}{2} arctan (\frac{1}{3} x))

Add a constant to the solution

= \frac{4}{15} ln tan (\frac{1}{2} arctan (\frac{1}{3} x)) + C

What is the integral of 4/(5xsqrt(x^2+9)) ?
The integral of 4/(5xsqrt(x^2+9)) is 4/15 ln|tan(1/2 arctan(1/3 x))|+C