What is the integral of 1/(usqrt(16u^2-9)) ?

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

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integral of 1/(usqrt(16u^2-9))

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

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

Solution steps

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

Apply Trigonometric Substitution

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

= \int \frac{1}{3} d v

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 v

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

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