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

The integral of (5x)/(9x^2-4) is 5/18 ln|9x^2-4|+C

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integral of (5x)/(9x^2-4)

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

\frac{5}{18} ln 9 x^{2} - 4 + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{18 u} d u

= 5 \cdot \int \frac{1}{18 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{18} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{18} ln ∣ u ∣

Substitute back

u = 9 x^{2} - 4

= 5 \cdot \frac{1}{18} ln 9 x^{2} - 4

Simplify

5 \cdot \frac{1}{18} ln 9 x^{2} - 4 : \frac{5}{18} ln 9 x^{2} - 4

= \frac{5}{18} ln 9 x^{2} - 4

Add a constant to the solution

= \frac{5}{18} ln 9 x^{2} - 4 + C

n = 0 \sum \infty e^{\frac{- n}{2}}

(x + y) d x + (x) d y = 0

\int (2 - sin (\frac{t}{4}))^{2} cos (\frac{t}{4}) d t

\frac{d y}{d x} = (cos (2 x))

y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

What is the integral of (5x)/(9x^2-4) ?
The integral of (5x)/(9x^2-4) is 5/18 ln|9x^2-4|+C