What is the integral of 1/(x^2sqrt(13-x^2)) ?

The integral of 1/(x^2sqrt(13-x^2)) is -(sqrt(13-x^2))/(13x)+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(13-x^2))

\int \frac{1}{x ^{2} 13 - x ^{2}} d x

- \frac{13 - x ^{2}}{13 x} + C

Solution steps

\int \frac{1}{x ^{2} 13 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{13 sin ( u ) cos ( u )} d u

= \int \frac{cot ( u )}{13 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}{13} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{13} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{13} (- cot (u))

Substitute back

u = arcsin (\frac{1}{13} x)

= \frac{1}{13} (- cot (arcsin (\frac{1}{13} x)))

Simplify

\frac{1}{13} (- cot (arcsin (\frac{1}{13} x))) : - \frac{13 - x ^{2}}{13 x}

= - \frac{13 - x ^{2}}{13 x}

Add a constant to the solution

= - \frac{13 - x ^{2}}{13 x} + C

\int \frac{2 ^{x}}{1 - 4 ^{x}} d x

\int \frac{2 x}{x ^{2} + k} d x

\int \frac{1}{( x ^{2} - 1 ) ( x ^{2} - 4 )} d x

\int \frac{x e ^{x}}{e ^{x} + 1} d x

What is the integral of 1/(x^2sqrt(13-x^2)) ?
The integral of 1/(x^2sqrt(13-x^2)) is -(sqrt(13-x^2))/(13x)+C