What is the integral of 1/(sqrt(4+25x^2)) ?

The integral of 1/(sqrt(4+25x^2)) is 1/5 ln((|5x+sqrt(4+25x^2)|)/2)+C

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integral of 1/(sqrt(4+25x^2))

\int \frac{1}{4 + 25 x ^{2}} d x

\frac{1}{5} ln (\frac{5 x + 4 + 25 x ^{2}}{2}) + C

Solution steps

\int \frac{1}{4 + 25 x ^{2}} 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 (\frac{5}{2} x)

= \frac{1}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))

Simplify

\frac{1}{5} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x)) : \frac{1}{5} ln (\frac{5 x + 4 + 25 x ^{2}}{2})

= \frac{1}{5} ln (\frac{5 x + 4 + 25 x ^{2}}{2})

Add a constant to the solution

= \frac{1}{5} ln (\frac{5 x + 4 + 25 x ^{2}}{2}) + C

\int \frac{x}{e ^{8 x}} d x

\int x^{2} a^{x} d x

\int \frac{ln ( 2 x )}{x ln ( 4 x )} d x

\int \frac{ln ( ln ( 3 x ) )}{x ln ( 3 x )} d x

2 x^{2}

What is the integral of 1/(sqrt(4+25x^2)) ?
The integral of 1/(sqrt(4+25x^2)) is 1/5 ln((|5x+sqrt(4+25x^2)|)/2)+C