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

The integral of 1/(x^2sqrt(9x^2-1)) is (sqrt(9x^2-1))/x+C

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

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

\frac{9 x ^{2} - 1}{x} + C

Solution steps

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

Apply Integral Substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Use the common integral:

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

= 4 \cdot \frac{3}{4} sin (v)

Substitute back

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

Simplify

4 \cdot \frac{3}{4} sin (arcsec (\frac{3}{2} \cdot 2 x)) : \frac{9 x ^{2} - 1}{x}

= \frac{9 x ^{2} - 1}{x}

Add a constant to the solution

= \frac{9 x ^{2} - 1}{x} + C

\int tan^{3} (11 π x) d x

\frac{d}{d x} (\frac{1}{4} sin (2 x))

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

\frac{d y}{d x} + 5 y = 20

y^{^{''}} + y = cot^{2} (x)

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