What is the integral of 1/(sqrt(x^2a^2)) ?

The integral of 1/(sqrt(x^2a^2)) is 1/a ln|x|+C

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integral of 1/(sqrt(x^2a^2))

\int \frac{1}{x ^{2} a ^{2}} d x

\frac{1}{a} ln ∣ x ∣ + C

Solution steps

\int \frac{1}{x ^{2} a ^{2}} d x

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x^{2} a^{2} = x^{2} a^{2}

= \int \frac{1}{x ^{2} a ^{2}} d x

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int \frac{1}{x ^{2}} d x

x^{2} = (x),

assuming

x \geq 0

= \frac{1}{a ^{2}} \cdot \int \frac{1}{x} d x

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{a ^{2}} ln ∣ x ∣

Apply radical rule:

assuming

a \geq 0

a^{2} = a

= \frac{1}{a} ln ∣ x ∣

Add a constant to the solution

= \frac{1}{a} ln ∣ x ∣ + C

n = 0 \sum \infty \frac{n + 1}{n ^{2}}

\frac{d}{d x} (2 a e^{2 x})

\int 14 x d x

f (x) = x (3 x - 12)^{4}

y = x^{2}, [0, 1]

What is the integral of 1/(sqrt(x^2a^2)) ?
The integral of 1/(sqrt(x^2a^2)) is 1/a ln|x|+C