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

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

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

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

\frac{b}{a} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int sec (u) d u

= \frac{b}{a} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{b}{a} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{1}{a} x)

= \frac{b}{a} ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x))

Simplify

\frac{b}{a} ln tan (arctan (\frac{1}{a} x)) + sec (arctan (\frac{1}{a} x)) : \frac{b}{a} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣})

= \frac{b}{a} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣})

Add a constant to the solution

= \frac{b}{a} ln (\frac{x + a ^{2} + x ^{2}}{∣ a ∣}) + C

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