What is the integral of 1/(x^2sqrt(7-16x^2)) ?

The integral of 1/(x^2sqrt(7-16x^2)) is -(sqrt(7-16x^2))/(7x)+C

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integral of 1/(x^2sqrt(7-16x^2))

\int \frac{1}{x ^{2} 7 - 16 x ^{2}} d x

- \frac{7 - 16 x ^{2}}{7 x} + C

Solution steps

\int \frac{1}{x ^{2} 7 - 16 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{4}{7 sin ^{2} ( u )} d u

= \int \frac{4}{7 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}{7} \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}{7} (- cot (u))

Substitute back

u = arcsin (\frac{4}{7} x)

= \frac{4}{7} (- cot (arcsin (\frac{4}{7} x)))

Simplify

\frac{4}{7} (- cot (arcsin (\frac{4}{7} x))) : - \frac{7 - 16 x ^{2}}{7 x}

= - \frac{7 - 16 x ^{2}}{7 x}

Add a constant to the solution

= - \frac{7 - 16 x ^{2}}{7 x} + C

\frac{\partial}{\partial y} (1 - x^{2} - y^{2})

\int \frac{1}{5 x - 2} d x

\int \frac{1}{u ^{4}} d u

\int 3 e^{x} d x

9 tan^{2} (7 x)

What is the integral of 1/(x^2sqrt(7-16x^2)) ?
The integral of 1/(x^2sqrt(7-16x^2)) is -(sqrt(7-16x^2))/(7x)+C