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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

\int \frac{1}{625 + 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}{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

\int x^{4} \cdot e^{12 x} d x

2 x y^{^{'}} + y = 2 x

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

y = x^{2} + 5 x + 3, at x = 1

\frac{d}{d x} (3 (5 - 7 x)^{5})

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