What is the integral of 1/(x^2sqrt(5-x^2)) ?

The integral of 1/(x^2sqrt(5-x^2)) is -(sqrt(5-x^2))/(5x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(x^2sqrt(5-x^2))

\int \frac{1}{x ^{2} 5 - x ^{2}} d x

- \frac{5 - x ^{2}}{5 x} + C

Solution steps

\int \frac{1}{x ^{2} 5 - x ^{2}} d x

Apply Trigonometric Substitution

= \int \frac{cot ( u )}{5 sin ( u ) cos ( u )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{5} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{5} (- cot (u))

Substitute back

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

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

Simplify

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

= - \frac{5 - x ^{2}}{5 x}

Add a constant to the solution

= - \frac{5 - x ^{2}}{5 x} + C

What is the integral of 1/(x^2sqrt(5-x^2)) ?
The integral of 1/(x^2sqrt(5-x^2)) is -(sqrt(5-x^2))/(5x)+C