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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

arctan (- 1 + x^{2}) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= arctan (u)

Substitute back

u = - 1 + x^{2}

= arctan (- 1 + x^{2})

Add a constant to the solution

= arctan (- 1 + x^{2}) + C

\int_{2}^{4} 4 x d x

x \to \frac{5 π}{2} + lim (sec (x))

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

\frac{\partial}{\partial y} (x ln (y))

f (x) = - 8

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