What is the integral of 1/(x^2sqrt(36x^2-1)) ?

The integral of 1/(x^2sqrt(36x^2-1)) is (sqrt(36x^2-1))/x+C

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integral of 1/(x^2sqrt(36x^2-1))

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

\frac{36 x ^{2} - 1}{x} + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Integral Substitution

= 2 \cdot \int \frac{4}{v ^{2} 9 v ^{2} - 4} d v

Take the constant out:

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

= 2 \cdot 4 \cdot \int \frac{1}{v ^{2} 9 v ^{2} - 4} d v

Apply Trigonometric Substitution

: \int \frac{3}{4 sec ( w )} d w

= 2 \cdot 4 \cdot \int \frac{3}{4 sec ( w )} d w

Take the constant out:

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

= 2 \cdot 4 \cdot \frac{3}{4} \cdot \int \frac{1}{sec ( w )} d w

Rewrite using trig identities

= 2 \cdot 4 \cdot \frac{3}{4} \cdot \int cos (w) d w

Use the common integral:

\int cos (w) d w = sin (w)

= 2 \cdot 4 \cdot \frac{3}{4} sin (w)

Substitute back

= 2 \cdot 4 \cdot \frac{3}{4} sin (arcsec (\frac{3}{2} \cdot 2 \cdot 2 x))

Simplify

2 \cdot 4 \cdot \frac{3}{4} sin (arcsec (\frac{3}{2} \cdot 2 \cdot 2 x)) : \frac{36 x ^{2} - 1}{x}

= \frac{36 x ^{2} - 1}{x}

Add a constant to the solution

= \frac{36 x ^{2} - 1}{x} + C

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What is the integral of 1/(x^2sqrt(36x^2-1)) ?
The integral of 1/(x^2sqrt(36x^2-1)) is (sqrt(36x^2-1))/x+C