What is the integral of 4/(xsqrt(x^2-36)) ?

The integral of 4/(xsqrt(x^2-36)) is 2/3 arcsec(1/6 x)+C

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integral of 4/(xsqrt(x^2-36))

\int \frac{4}{x x ^{2} - 36} d x

\frac{2}{3} arcsec (\frac{1}{6} x) + C

Solution steps

\int \frac{4}{x x ^{2} - 36} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{x x ^{2} - 36} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

= 4 \cdot \int \frac{1}{6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 4 \cdot \frac{1}{6} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{6} x)

= 4 \cdot \frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} x)

Simplify

4 \cdot \frac{1}{6} \cdot 1 \cdot arcsec (\frac{1}{6} x) : \frac{2}{3} arcsec (\frac{1}{6} x)

= \frac{2}{3} arcsec (\frac{1}{6} x)

Add a constant to the solution

= \frac{2}{3} arcsec (\frac{1}{6} x) + C

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What is the integral of 4/(xsqrt(x^2-36)) ?
The integral of 4/(xsqrt(x^2-36)) is 2/3 arcsec(1/6 x)+C