What is the integral of 1/(sqrt(x^2+169)) ?

The integral of 1/(sqrt(x^2+169)) is ln(1/13 |x+sqrt(169+x^2)|)+C

Popular >

integral of 1/(sqrt(x^2+169))

\int \frac{1}{x ^{2} + 169} d x

ln (\frac{1}{13} x + 169 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 169} 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}{13} x)

= ln tan (arctan (\frac{1}{13} x)) + sec (arctan (\frac{1}{13} x))

Simplify

ln tan (arctan (\frac{1}{13} x)) + sec (arctan (\frac{1}{13} x)) : ln (\frac{1}{13} x + 169 + x^{2})

= ln (\frac{1}{13} x + 169 + x^{2})

Add a constant to the solution

= ln (\frac{1}{13} x + 169 + x^{2}) + C

(2 x + 1) (\frac{d y}{d x}) + y = (2 x + 1)^{\frac{3}{2}}

y^{^{''}} + 3 y = - 48 x^{2} e^{3 x}

f (x) = arcsinh (x)

e^{(- 4 s)}

\int - 510 x^{2} + 3414 x + 1998 d x

What is the integral of 1/(sqrt(x^2+169)) ?
The integral of 1/(sqrt(x^2+169)) is ln(1/13 |x+sqrt(169+x^2)|)+C