What is the integral of 1/(x^2sqrt(225-x^2)) ?

The integral of 1/(x^2sqrt(225-x^2)) is -(sqrt(225-x^2))/(225x)+C

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integral of 1/(x^2sqrt(225-x^2))

\int \frac{1}{x ^{2} 225 - x ^{2}} d x

- \frac{225 - x ^{2}}{225 x} + C

Solution steps

\int \frac{1}{x ^{2} 225 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{225 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{225 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{225} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{225} \cdot \int csc^{2} (u) d u

Use the common integral:

\int csc^{2} (u) d u = - cot (u)

= \frac{1}{225} (- cot (u))

Substitute back

u = arcsin (\frac{1}{15} x)

= \frac{1}{225} (- cot (arcsin (\frac{1}{15} x)))

Simplify

\frac{1}{225} (- cot (arcsin (\frac{1}{15} x))) : - \frac{225 - x ^{2}}{225 x}

= - \frac{225 - x ^{2}}{225 x}

Add a constant to the solution

= - \frac{225 - x ^{2}}{225 x} + C

\frac{d y}{d x} = sin (6 x)

y^{^{'}} = 3 - 2 y

\int 7 tan^{2} (x) d x

6 e^{5 t} cos (2 t) - e^{7 t}

- 7 e^{- 3 x} - 6 e^{- 3 x} x

What is the integral of 1/(x^2sqrt(225-x^2)) ?
The integral of 1/(x^2sqrt(225-x^2)) is -(sqrt(225-x^2))/(225x)+C