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

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

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

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

ln \frac{x}{c} + \frac{c + x ^{2}}{c} + C

Solution steps

\int \frac{1}{x ^{2} + c} 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}{c} x)

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

Simplify

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

= ln \frac{x}{c} + \frac{c + x ^{2}}{c}

Add a constant to the solution

= ln \frac{x}{c} + \frac{c + x ^{2}}{c} + C

\frac{2}{( s - 4 ) ^{3}}

\int_{- π}^{π} sin (k x) cos (n x) d x

x^{^{'}} = a x - b x^{2}

x \to - 1 lim (\frac{f ( x )}{x + 1})

y^{^{'}} = \frac{- x y ^{2}}{2}

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