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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{5} d u

= \int \frac{sec ( u )}{5} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{5} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{5} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (5 x)

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{3}{4 + 6 n}

n = 1 \sum \infty \frac{2}{3 n + 5}

sin (x) d x + y d y = 0

\int x^{9} e^{x^{5}} d x

\frac{d}{d x} (\frac{a + b x + c x ^{2}}{x})

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