What is the integral of 1/(xsqrt(4^2-9)) ?

The integral of 1/(xsqrt(4^2-9)) is 1/(sqrt(7))ln|x|+C

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integral of 1/(xsqrt(4^2-9))

\int \frac{1}{x 4 ^{2} - 9} d x

\frac{1}{7} ln ∣ x ∣ + C

Solution steps

\int \frac{1}{x 4 ^{2} - 9} d x

Take the constant out:

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

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

Simplify

= \frac{1}{7} \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{7} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ x ∣ + C

\int \frac{1}{1 - \frac{1}{2} x} d x

\frac{x}{6}

(x tan (\frac{y}{x}) + y) \cdot d x - x \cdot d y = 0

\int \frac{1}{x} - \frac{1}{x + 6} d x

\int y 2 + 10 y - 25 y^{2} d y

What is the integral of 1/(xsqrt(4^2-9)) ?
The integral of 1/(xsqrt(4^2-9)) is 1/(sqrt(7))ln|x|+C