What is the integral of 1/(xsqrt(x^2-100)) ?

The integral of 1/(xsqrt(x^2-100)) is 1/10 arcsec(1/10 x)+C

Popular >

integral of 1/(xsqrt(x^2-100))

\int \frac{1}{x x ^{2} - 100} d x

\frac{1}{10} arcsec (\frac{1}{10} x) + C

Solution steps

\int \frac{1}{x x ^{2} - 100} d x

Apply Trigonometric Substitution

: \int \frac{1}{10} d u

= \int \frac{1}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{10} \cdot 1 \cdot u

Substitute back

u = arcsec (\frac{1}{10} x)

= \frac{1}{10} \cdot 1 \cdot arcsec (\frac{1}{10} x)

Simplify

= \frac{1}{10} arcsec (\frac{1}{10} x)

Add a constant to the solution

= \frac{1}{10} arcsec (\frac{1}{10} x) + C

\frac{\partial}{\partial y} (e^{x y z^{7}} y z^{7})

\frac{d v}{d t} = k (250 - v)

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

\frac{d}{d x} (\frac{1 + x}{x})

\int \frac{2 x - 3}{1 - x ^{2}} d x

What is the integral of 1/(xsqrt(x^2-100)) ?
The integral of 1/(xsqrt(x^2-100)) is 1/10 arcsec(1/10 x)+C