What is the integral of 1/(x^2sqrt(196-x^2)) ?

The integral of 1/(x^2sqrt(196-x^2)) is -(sqrt(196-x^2))/(196x)+C

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integral of 1/(x^2sqrt(196-x^2))

\int \frac{1}{x ^{2} 196 - x ^{2}} d x

- \frac{196 - x ^{2}}{196 x} + C

Solution steps

\int \frac{1}{x ^{2} 196 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{196 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{196 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}{196} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{196} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{196} (- cot (u))

Substitute back

u = arcsin (\frac{1}{14} x)

= \frac{1}{196} (- cot (arcsin (\frac{1}{14} x)))

Simplify

\frac{1}{196} (- cot (arcsin (\frac{1}{14} x))) : - \frac{196 - x ^{2}}{196 x}

= - \frac{196 - x ^{2}}{196 x}

Add a constant to the solution

= - \frac{196 - x ^{2}}{196 x} + C

(x - 2) (x + 1) (x + 6)

y = 4 + 8 x^{2}, (0, 4)

tan (4 x^{2} - 3 x - 4)

x \to 2 - lim (\frac{x}{x ^{2} + 4})

\int \frac{3}{4} + \frac{4}{5} x^{2} - \frac{5}{6} x^{3} d x

What is the integral of 1/(x^2sqrt(196-x^2)) ?
The integral of 1/(x^2sqrt(196-x^2)) is -(sqrt(196-x^2))/(196x)+C