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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

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

= ln ∣ tan (v) + sec (v) ∣

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

(9, 1), (- 1, 1)

\int - \frac{2}{x ^{3}} d x

\int 4 - x d x

\frac{d}{d x} (\frac{8}{x ^{2} + 1})

x (x^{2} + 1)^{3}

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