What is the integral of 1/(x^2sqrt(25-x^2)) ?

The integral of 1/(x^2sqrt(25-x^2)) is -(sqrt(25-x^2))/(25x)+C

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integral of 1/(x^2sqrt(25-x^2))

\int \frac{1}{x ^{2} 25 - x ^{2}} d x

- \frac{25 - x ^{2}}{25 x} + C

Solution steps

\int \frac{1}{x ^{2} 25 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{25 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{25 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}{25} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{25} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{25} (- cot (u))

Substitute back

u = arcsin (\frac{1}{5} x)

= \frac{1}{25} (- cot (arcsin (\frac{1}{5} x)))

Simplify

\frac{1}{25} (- cot (arcsin (\frac{1}{5} x))) : - \frac{25 - x ^{2}}{25 x}

= - \frac{25 - x ^{2}}{25 x}

Add a constant to the solution

= - \frac{25 - x ^{2}}{25 x} + C

x \cdot (ln (x))^{2}

\int sin (x) + cos (y) d x

\int \frac{9 - x ^{2}}{( x ^{2} )} d x

\int 2 cos (t) sin (t) d t

\int 4 x - 3 d x

What is the integral of 1/(x^2sqrt(25-x^2)) ?
The integral of 1/(x^2sqrt(25-x^2)) is -(sqrt(25-x^2))/(25x)+C