What is the integral of 1/(sqrt(x^2+25)) ?

The integral of 1/(sqrt(x^2+25)) is ln(1/5 |x+sqrt(25+x^2)|)+C

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integral of 1/(sqrt(x^2+25))

\int \frac{1}{x ^{2} + 25} d x

ln (\frac{1}{5} x + 25 + x^{2}) + C

Solution steps

\int \frac{1}{x ^{2} + 25} 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}{5} x)

= ln tan (arctan (\frac{1}{5} x)) + sec (arctan (\frac{1}{5} x))

Simplify

ln tan (arctan (\frac{1}{5} x)) + sec (arctan (\frac{1}{5} x)) : ln (\frac{1}{5} x + 25 + x^{2})

= ln (\frac{1}{5} x + 25 + x^{2})

Add a constant to the solution

= ln (\frac{1}{5} x + 25 + x^{2}) + C

\int x (1 + x)^{4} d x

y^{^{'}} = t - y t

\frac{d y}{d x}, y = 2 x + 1

\int sin (2 \cdot x) d x

\int 15 - 14 x - x^{2} d x

What is the integral of 1/(sqrt(x^2+25)) ?
The integral of 1/(sqrt(x^2+25)) is ln(1/5 |x+sqrt(25+x^2)|)+C