What is the integral of 1/(x^2sqrt(9-16x^2)) ?

The integral of 1/(x^2sqrt(9-16x^2)) is -(sqrt(9-16x^2))/(9x)+C

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integral of 1/(x^2sqrt(9-16x^2))

\int \frac{1}{x ^{2} 9 - 16 x ^{2}} d x

- \frac{9 - 16 x ^{2}}{9 x} + C

Solution steps

\int \frac{1}{x ^{2} 9 - 16 x ^{2}} d x

Apply Trigonometric Substitution

= \int \frac{4}{9 sin ^{2} ( u )} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int \frac{1}{sin ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{sin ^{2} ( u )} d u = - cot (u)

= \frac{4}{9} (- cot (u))

Substitute back

u = arcsin (\frac{4}{3} x)

= \frac{4}{9} (- cot (arcsin (\frac{4}{3} x)))

Simplify

\frac{4}{9} (- cot (arcsin (\frac{4}{3} x))) : - \frac{9 - 16 x ^{2}}{9 x}

= - \frac{9 - 16 x ^{2}}{9 x}

Add a constant to the solution

= - \frac{9 - 16 x ^{2}}{9 x} + C

What is the integral of 1/(x^2sqrt(9-16x^2)) ?
The integral of 1/(x^2sqrt(9-16x^2)) is -(sqrt(9-16x^2))/(9x)+C