What is the integral of 1/(sqrt(x^2+625)) ?

The integral of 1/(sqrt(x^2+625)) is ln(1/25 |x+sqrt(625+x^2)|)+C

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integral of 1/(sqrt(x^2+625))

\int \frac{1}{x ^{2} + 625} d x

ln (\frac{1}{25} x + 625 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 625} 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}{25} x)

= ln tan (arctan (\frac{1}{25} x)) + sec (arctan (\frac{1}{25} x))

Simplify

ln tan (arctan (\frac{1}{25} x)) + sec (arctan (\frac{1}{25} x)) : ln (\frac{1}{25} x + 625 + x^{2})

= ln (\frac{1}{25} x + 625 + x^{2})

Add a constant to the solution

= ln (\frac{1}{25} x + 625 + x^{2}) + C

x \to 0 lim (\frac{∣ x ∣}{x ^{3} - 5})

y = 8 x^{2} - 5 x^{11}

\int (x - 2) e^{x + 3} d x

\frac{\partial}{\partial x} (4 x^{2} - y^{2} + 2 y)

y = \frac{x}{3}, y = x - 2

What is the integral of 1/(sqrt(x^2+625)) ?
The integral of 1/(sqrt(x^2+625)) is ln(1/25 |x+sqrt(625+x^2)|)+C