What is the integral of 1/(x^2(sqrt(x^2-16))) ?

The integral of 1/(x^2(sqrt(x^2-16))) is (sqrt(x^2-16))/(16x)+C

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integral of 1/(x^2(sqrt(x^2-16)))

\int \frac{1}{x ^{2} ( x ^{2} - 16 )} d x

\frac{x ^{2} - 16}{16 x} + C

Solution steps

\int \frac{1}{x ^{2} x ^{2} - 16} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 sec ( u )} d u

= \int \frac{1}{16 sec ( u )} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{16} \cdot \int cos (u) d u

Use the common integral:

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

= \frac{1}{16} sin (u)

Substitute back

u = arcsec (\frac{1}{4} x)

= \frac{1}{16} sin (arcsec (\frac{1}{4} x))

Simplify

\frac{1}{16} sin (arcsec (\frac{1}{4} x)) : \frac{x ^{2} - 16}{16 x}

= \frac{x ^{2} - 16}{16 x}

Add a constant to the solution

= \frac{x ^{2} - 16}{16 x} + C

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\frac{\partial}{\partial y} (e^{- y} cos (π x))

\int (x^{2} + e^{x} - sin (x)) d x

(3 x)^{^{'}}

What is the integral of 1/(x^2(sqrt(x^2-16))) ?
The integral of 1/(x^2(sqrt(x^2-16))) is (sqrt(x^2-16))/(16x)+C