What is the integral of 1/(x^2sqrt(49-x^2)) ?

The integral of 1/(x^2sqrt(49-x^2)) is -(sqrt(49-x^2))/(49x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2sqrt(49-x^2))

\int \frac{1}{x ^{2} 49 - x ^{2}} d x

- \frac{49 - x ^{2}}{49 x} + C

Solution steps

\int \frac{1}{x ^{2} 49 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{49 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{49 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}{49} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{49} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{49} (- cot (u))

Substitute back

u = arcsin (\frac{1}{7} x)

= \frac{1}{49} (- cot (arcsin (\frac{1}{7} x)))

Simplify

\frac{1}{49} (- cot (arcsin (\frac{1}{7} x))) : - \frac{49 - x ^{2}}{49 x}

= - \frac{49 - x ^{2}}{49 x}

Add a constant to the solution

= - \frac{49 - x ^{2}}{49 x} + C

f (x) = x^{2}, at x = 0

\int 7 sin^{2} (t) cos^{4} (t) d t

\frac{( 2 t e ^{- 3 t} )}{5} + 3 t \cdot sin (t)

(13 + x^{2}) y^{^{'}} - 2 x y = 0

\frac{\partial}{\partial x} (y x + 2 x + g (y))

What is the integral of 1/(x^2sqrt(49-x^2)) ?
The integral of 1/(x^2sqrt(49-x^2)) is -(sqrt(49-x^2))/(49x)+C