What is the integral of 1/(xsqrt(16x^2-9)) ?

The integral of 1/(xsqrt(16x^2-9)) is 1/3 arcsec(4/3 x)+C

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integral of 1/(xsqrt(16x^2-9))

\int \frac{1}{x 16 x ^{2} - 9} d x

\frac{1}{3} arcsec (\frac{4}{3} x) + C

Solution steps

\int \frac{1}{x 16 x ^{2} - 9} d x

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

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

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{4}{3} x)

= \frac{1}{3} \cdot 1 \cdot arcsec (\frac{4}{3} x)

Simplify

= \frac{1}{3} arcsec (\frac{4}{3} x)

Add a constant to the solution

= \frac{1}{3} arcsec (\frac{4}{3} x) + C

\frac{\partial}{\partial x} (\frac{7}{x ^{3}})

\frac{d}{d x} (\frac{π x}{2})

\int (sin^{3} (x) cos (x)) d x

\int e^{- \frac{2 r}{a}} r^{2} d r

f (x) = x^{3}, at - 5

What is the integral of 1/(xsqrt(16x^2-9)) ?
The integral of 1/(xsqrt(16x^2-9)) is 1/3 arcsec(4/3 x)+C