What is the integral of 1/(sqrt(324+x^2)) ?

The integral of 1/(sqrt(324+x^2)) is ln(1/18 |x+sqrt(324+x^2)|)+C

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integral of 1/(sqrt(324+x^2))

\int \frac{1}{324 + x ^{2}} d x

ln (\frac{1}{18} x + 324 + x^{2}) + C

Solution steps

\int \frac{1}{324 + x ^{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}{18} x)

= ln tan (arctan (\frac{1}{18} x)) + sec (arctan (\frac{1}{18} x))

Simplify

ln tan (arctan (\frac{1}{18} x)) + sec (arctan (\frac{1}{18} x)) : ln (\frac{1}{18} x + 324 + x^{2})

= ln (\frac{1}{18} x + 324 + x^{2})

Add a constant to the solution

= ln (\frac{1}{18} x + 324 + x^{2}) + C

\frac{\partial}{\partial y} (e^{x} sin (y))

\int \frac{ln ( 5 x )}{x ^{3}} d x

y = 4 x, y = x^{2}

\int_{0}^{1} t d t

h \to 0 lim (5 h + 56)

What is the integral of 1/(sqrt(324+x^2)) ?
The integral of 1/(sqrt(324+x^2)) is ln(1/18 |x+sqrt(324+x^2)|)+C