What is the integral of 1/(x^2sqrt(x^2-169)) ?

The integral of 1/(x^2sqrt(x^2-169)) is (sqrt(x^2-169))/(169x)+C

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integral of 1/(x^2sqrt(x^2-169))

\int \frac{1}{x ^{2} x ^{2} - 169} d x

\frac{x ^{2} - 169}{169 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} - 169} d x

Apply Trigonometric Substitution

: \int \frac{1}{169 sec ( u )} d u

= \int \frac{1}{169 sec ( u )} d u

Take the constant out:

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

= \frac{1}{169} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{169} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{169} sin (u)

Substitute back

u = arcsec (\frac{1}{13} x)

= \frac{1}{169} sin (arcsec (\frac{1}{13} x))

Simplify

\frac{1}{169} sin (arcsec (\frac{1}{13} x)) : \frac{x ^{2} - 169}{169 x}

= \frac{x ^{2} - 169}{169 x}

Add a constant to the solution

= \frac{x ^{2} - 169}{169 x} + C

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

y^{^{''}} + 4 y^{^{'}} + 20 y = 80

y = x, y = x - 2, y = 0

\int \frac{1}{( x ^{2} + 81 ) ^{2}} d x

\int 4 cos^{2} (θ) d θ

What is the integral of 1/(x^2sqrt(x^2-169)) ?
The integral of 1/(x^2sqrt(x^2-169)) is (sqrt(x^2-169))/(169x)+C