What is the integral of 1/(sqrt(x^2+8)) ?

The integral of 1/(sqrt(x^2+8)) is ln|x+sqrt(8+x^2)|-3/2 ln(2)+C

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integral of 1/(sqrt(x^2+8))

\int \frac{1}{x ^{2} + 8} d x

ln x + 8 + x^{2} - \frac{3}{2} ln (2) + C

Solution steps

\int \frac{1}{x ^{2} + 8} d x

Apply Trigonometric Substitution

= \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}{2 2} x)

= ln tan (arctan (\frac{1}{2 2} x)) + sec (arctan (\frac{1}{2 2} x))

Simplify

ln tan (arctan (\frac{1}{2 2} x)) + sec (arctan (\frac{1}{2 2} x)) : ln x + 8 + x^{2} - \frac{3}{2} ln (2)

= ln x + 8 + x^{2} - \frac{3}{2} ln (2)

Add a constant to the solution

= ln x + 8 + x^{2} - \frac{3}{2} ln (2) + C

What is the integral of 1/(sqrt(x^2+8)) ?
The integral of 1/(sqrt(x^2+8)) is ln|x+sqrt(8+x^2)|-3/2 ln(2)+C