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

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

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

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

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

Solution steps

\int \frac{1}{x 25 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{5}{3} x)

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

Simplify

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

Add a constant to the solution

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

\int \frac{ln ( 1 + x ) - ln ( x )}{x ( 1 + x )} d x

\int sin (\frac{1}{3}) x sec^{2} (\frac{1}{3}) x d x

\int 9 t^{4} + 16 t^{2} d t

\int t^{2} 1 + \frac{1}{t ^{2}} d t

\int e^{t x} x e^{- x} d x

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