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

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

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

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

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

Solution steps

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

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{2} 1

\int (3 x^{5} - 4)^{2} \cdot 5 x^{4} d x

y^{^{''}} + 2 \cdot 2.5 \cdot 4 \cdot y^{^{'}} + 16 y = 0

\int \frac{1}{3 + 5 cos ( x )} d x

\frac{d}{d x} (x^{3} e^{5 x})

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