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

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

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

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

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

Solution steps

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

\int (8 x + 3 sin (x))^{2} d x

\int (\frac{1}{4 - 25 x ^{2}}) d x

\frac{14 x + 4 x ^{2}}{( 7 + 4 x ) ^{2}}

y^{^{'}} = \frac{6 x ^{3} + y ^{3}}{x y ^{2}}

\int \frac{( x ^{2} + 4 )}{3 x ^{3} + 4 x ^{2} - 4 x} d x

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