What is the integral of 1/(sqrt(x^2+81)) ?

The integral of 1/(sqrt(x^2+81)) is ln(1/9 |x+sqrt(81+x^2)|)+C

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integral of 1/(sqrt(x^2+81))

\int \frac{1}{x ^{2} + 81} d x

ln (\frac{1}{9} x + 81 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 81} 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}{9} x)

= ln tan (arctan (\frac{1}{9} x)) + sec (arctan (\frac{1}{9} x))

Simplify

ln tan (arctan (\frac{1}{9} x)) + sec (arctan (\frac{1}{9} x)) : ln (\frac{1}{9} x + 81 + x^{2})

= ln (\frac{1}{9} x + 81 + x^{2})

Add a constant to the solution

= ln (\frac{1}{9} x + 81 + x^{2}) + C

\int_{1}^{x} 1 + 2 t d t

\frac{d}{d x} (\frac{4}{3} \cdot π \cdot x^{3})

\int x^{3} 5 x d x

x \to 0 lim (\frac{cos ( 6 x )}{1})

x \to 3 lim (\frac{2 x - 6}{x - 3})

What is the integral of 1/(sqrt(x^2+81)) ?
The integral of 1/(sqrt(x^2+81)) is ln(1/9 |x+sqrt(81+x^2)|)+C