What is the integral of 1/(xsqrt(9x^2-25)) ?

The integral of 1/(xsqrt(9x^2-25)) is 1/5 arcsec(3/5 x)+C

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integral of 1/(xsqrt(9x^2-25))

\int \frac{1}{x 9 x ^{2} - 25} d x

\frac{1}{5} arcsec (\frac{3}{5} x) + C

Solution steps

\int \frac{1}{x 9 x ^{2} - 25} d x

Apply Trigonometric Substitution

: \int \frac{1}{5} d u

= \int \frac{1}{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 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot u

Substitute back

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

= \frac{1}{5} \cdot 1 \cdot arcsec (\frac{3}{5} x)

Simplify

= \frac{1}{5} arcsec (\frac{3}{5} x)

Add a constant to the solution

= \frac{1}{5} arcsec (\frac{3}{5} x) + C

\frac{d y}{d x} = \frac{4}{x ^{2} e ^{2 y}}

\int sin (x) d x

\frac{\partial}{\partial x} (6 x^{7} y^{4} + 2 x^{6} y^{8})

4 y^{3} + 7 x^{8} = 5 y + 6 x, (1, 1)

n = 1 \sum \infty (\frac{5}{2})^{n}

What is the integral of 1/(xsqrt(9x^2-25)) ?
The integral of 1/(xsqrt(9x^2-25)) is 1/5 arcsec(3/5 x)+C