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

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

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

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

\frac{1}{b} ln (\frac{b x + a ^{2} + b ^{2} x ^{2}}{∣ a ∣}) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{b} d u

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

Take the constant out:

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

= \frac{1}{b} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{b} ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

= \frac{1}{b} ln (\frac{b x + a ^{2} + b ^{2} x ^{2}}{∣ a ∣})

Add a constant to the solution

= \frac{1}{b} ln (\frac{b x + a ^{2} + b ^{2} x ^{2}}{∣ a ∣}) + C

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