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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec (u) d u

= \int sec (u) d u

Use the common integral:

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

= ln ∣ tan (u) + sec (u) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 5 x - x^{2}, at x = 1

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

\int (x^{2} + 3 x - 1) \cdot e^{2 x} d x

\frac{\partial}{\partial x} (5 x^{2 y})

\frac{d}{d x} (ln (csc (x) - cot (x)))

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