What is the integral of 1/(x^2sqrt(400-x^2)) ?

The integral of 1/(x^2sqrt(400-x^2)) is -(sqrt(400-x^2))/(400x)+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(400-x^2))

\int \frac{1}{x ^{2} 400 - x ^{2}} d x

- \frac{400 - x ^{2}}{400 x} + C

Solution steps

\int \frac{1}{x ^{2} 400 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{400 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{400 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}{400} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{400} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{400} (- cot (u))

Substitute back

u = arcsin (\frac{1}{20} x)

= \frac{1}{400} (- cot (arcsin (\frac{1}{20} x)))

Simplify

\frac{1}{400} (- cot (arcsin (\frac{1}{20} x))) : - \frac{400 - x ^{2}}{400 x}

= - \frac{400 - x ^{2}}{400 x}

Add a constant to the solution

= - \frac{400 - x ^{2}}{400 x} + C

\frac{d}{d x} ((2 x^{3} + 4) (x^{4} - 4 x))

\int_{0}^{1} \frac{5}{1 - x ^{2}} d x

y = (x + 5) (x^{3} - 1) (x^{2} - x + 3)

\frac{d}{d x} (2 e^{7 x})

\int \frac{1}{5 - 4 sin ( x ) + 3 cos ( x )} d x

What is the integral of 1/(x^2sqrt(400-x^2)) ?
The integral of 1/(x^2sqrt(400-x^2)) is -(sqrt(400-x^2))/(400x)+C