intergate

[expscv]

..._
_/ 4/ (x^2-2x-1) dx

thx

[himanshu121]

..._
_/ 4/ (x^2-2x-1) dx

thx

\int \frac{4}{x^2-2x-1} dx = \int \frac{4}{(x-1)^2-2}dx

U can apply basic formula
\int \frac{1}{x^2-a^2}dx = \frac{1}{2a} \ln { \frac {x-a} {x+a}}

[expscv]

great thx, i forgot the formula =P

[expscv]

actually i never leanred that formula =(

my most silmmilar was \int \frac{1}{a^2+x^2} =\frac{1}{a}{tan^-1(\frac{x}{a})

[expscv]

latex is odd to use espically for fraction

1/x or \frac {1}{x}