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

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

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

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

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\int x sin (π) x d x

\frac{\partial}{\partial x} (4 x^{2} \cdot 3 x^{3} y^{2})

5 x + 1

x \to 0 + lim ((3 x)^{x})

\frac{d}{d x} (- 2 \cdot x)

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