What is the integral of 1/(xsqrt(9x^2-49)) ?

The integral of 1/(xsqrt(9x^2-49)) is 1/7 arcsec(3/7 x)+C

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integral of 1/(xsqrt(9x^2-49))

\int \frac{1}{x 9 x ^{2} - 49} d x

\frac{1}{7} arcsec (\frac{3}{7} x) + C

Solution steps

\int \frac{1}{x 9 x ^{2} - 49} d x

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= \int \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{3}{7} x)

= \frac{1}{7} \cdot 1 \cdot arcsec (\frac{3}{7} x)

Simplify

= \frac{1}{7} arcsec (\frac{3}{7} x)

Add a constant to the solution

= \frac{1}{7} arcsec (\frac{3}{7} x) + C

\frac{d}{d x} (\frac{1}{( 1 + x ) ^{3}})

4 y^{^{''}} + 16 y^{^{'}} + 19 y = 0

\frac{d}{d x} (arctan (\frac{d}{x}))

\int (x - 1) e^{x} d x

y^{^{''}} + 2 y^{^{'}} = (8 t - 7) sin (3 t)

What is the integral of 1/(xsqrt(9x^2-49)) ?
The integral of 1/(xsqrt(9x^2-49)) is 1/7 arcsec(3/7 x)+C