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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

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

f (x) = x^{x} (ln (x) + 1)

f (x) = x e^{4 x}

\frac{\partial}{\partial y} (sin (x^{2} - y^{2}))

cos (e^{x})

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