What is the integral of 1/(sqrt(x^2-0.01)) ?

The integral of 1/(sqrt(x^2-0.01)) is ln|sqrt(100x^2-1)+10x|+C

integral of 1/(sqrt(x^2-0.01))

\int \frac{1}{x ^{2} - 0.01} d x

ln 100 x^{2} - 1 + 10 x + C

Solution steps

\int \frac{1}{x ^{2} - 0.01} d x

Apply Trigonometric Substitution

= \int \frac{0.1 sec ( u )}{0.01} d u

Simplify

= \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= ln ∣ tan (u) + sec (u) ∣

Substitute back

u = arcsec (\frac{1}{0.1} x)

= ln tan (arcsec (\frac{1}{0.1} x)) + sec (arcsec (\frac{1}{0.1} x))

Simplify

ln tan (arcsec (\frac{1}{0.1} x)) + sec (arcsec (\frac{1}{0.1} x)) : ln 100 x^{2} - 1 + 10 x

= ln 100 x^{2} - 1 + 10 x

Add a constant to the solution

= ln 100 x^{2} - 1 + 10 x + C

\int \frac{x ^{3}}{1 + x} d x

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 4 t^{2}

\int_{- 2}^{4} ∣ - 2 x + 4 ∣ d x

y^{^{'}} - \frac{2}{x} y = 0

\frac{\partial}{\partial x} (\frac{7 x ^{2}}{( x - y ) ^{2}})

What is the integral of 1/(sqrt(x^2-0.01)) ?
The integral of 1/(sqrt(x^2-0.01)) is ln|sqrt(100x^2-1)+10x|+C