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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{13 sec ( u )} d u

= \int \frac{1}{13 sec ( 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{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{13} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{13} sin (u)

Substitute back

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

= \frac{1}{13} sin (arcsec (\frac{1}{13} x))

Simplify

\frac{1}{13} sin (arcsec (\frac{1}{13} x)) : \frac{x ^{2} - 13}{13 x}

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

Add a constant to the solution

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

\frac{d y}{d x} = 14 x e^{6} x (36 + y^{2})

\int 9 sin^{6} (x) d x

y^{^{'}} + y^{3} x + 9 y = 0

\int x^{5} + 8 x^{- 3} + 2 d x

\frac{d}{d x} (3 x^{3} + 5 x)

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