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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + \frac{x ^{2}}{2} + C

Solution steps

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

Multiply by the conjugate of

1 + x^{2} - x : \frac{1}{1 + x ^{2} + x}

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

Simplify

= \int 1 + x^{2} + x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 + x^{2} d x + \int x d x

\int 1 + x^{2} d x = \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2}

\int x d x = \frac{x ^{2}}{2}

= \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + \frac{x ^{2}}{2}

Add a constant to the solution

= \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + \frac{x ^{2}}{2} + C

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