What is the integral of 1/(sqrt(9x^2+5)) ?

The integral of 1/(sqrt(9x^2+5)) is 1/3 ln| 3/(sqrt(5))x+sqrt(1/5 (5+9x^2))|+C

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integral of 1/(sqrt(9x^2+5))

\int \frac{1}{9 x ^{2} + 5} d x

\frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 9 x^{2}) + C

Solution steps

\int \frac{1}{9 x ^{2} + 5} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{3} d u

= \int \frac{sec ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (u) d u

Use the common integral:

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

= \frac{1}{3} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arctan (\frac{3}{5} x)

= \frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x))

Simplify

\frac{1}{3} ln tan (arctan (\frac{3}{5} x)) + sec (arctan (\frac{3}{5} x)) : \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 9 x^{2})

= \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 9 x^{2})

Add a constant to the solution

= \frac{1}{3} ln \frac{3}{5} x + \frac{1}{5} (5 + 9 x^{2}) + C

144 y^{^{'''}} + 120 y^{^{''}} + 25 y^{^{'}} = 0

\int (4 x + 7) e^{x} d x

\int 30 x^{29} d x

\frac{d ^{2}}{d x ^{2}} (\frac{3 x + 2}{2 x - 5})

25 y^{^{''}} - 20 y^{^{'}} + 4 y = 0, y (5) = 0, y^{^{'}} (0) = - e^{2}

What is the integral of 1/(sqrt(9x^2+5)) ?
The integral of 1/(sqrt(9x^2+5)) is 1/3 ln| 3/(sqrt(5))x+sqrt(1/5 (5+9x^2))|+C