What is the integral of 1/(x^2sqrt(x^2-196)) ?

The integral of 1/(x^2sqrt(x^2-196)) is (sqrt(x^2-196))/(196x)+C

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integral of 1/(x^2sqrt(x^2-196))

\int \frac{1}{x ^{2} x ^{2} - 196} d x

\frac{x ^{2} - 196}{196 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} - 196} d x

Apply Trigonometric Substitution

: \int \frac{1}{196 sec ( u )} d u

= \int \frac{1}{196 sec ( u )} d u

Take the constant out:

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

= \frac{1}{196} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{196} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{196} sin (u)

Substitute back

u = arcsec (\frac{1}{14} x)

= \frac{1}{196} sin (arcsec (\frac{1}{14} x))

Simplify

\frac{1}{196} sin (arcsec (\frac{1}{14} x)) : \frac{x ^{2} - 196}{196 x}

= \frac{x ^{2} - 196}{196 x}

Add a constant to the solution

= \frac{x ^{2} - 196}{196 x} + C

\int \frac{e ^{x} ( 2 x ^{3} + x ^{2} - 4 x - 6 )}{6 x ^{4}} d x

t \to 0 lim (\frac{sin ( 2 t )}{t})

x \to 0 + lim (sin (x) \cdot ln (x))

4 y^{^{''}} - 9 y = 0

y = \frac{1}{t + 1}

What is the integral of 1/(x^2sqrt(x^2-196)) ?
The integral of 1/(x^2sqrt(x^2-196)) is (sqrt(x^2-196))/(196x)+C