Question

`int dx/(x (x^2 + 1))` बराबर है:

Options

• `log abs x - 1/2  log (x^2 + 1) + C`

• `log abs x + 1/2  log (x^2 + 1) + C`

• `- log abs x + 1/2  log (x^2 + 1) + C`

• `1/2  log abs x -  log (x^2 + 1) + C`

Answer 1

`log abs x - 1/2  log (x^2 + 1) + C`

स्पष्टीकरण:

माना `I = int dx/(x (x^2 + 1))`

`= int x/(x (x^2 + 1))  dx`

x² = t रखने पर,

2x dx = dt

`I = 1/2 int (2x  dx)/(x (x^2 + 1))`

`= 1/2 int dt/(t (t + 1))`

अब, `1/(t (t + 1)) = A/t + B/(t + 1)`

1 = A(t + 1) + Bt

t = 0 रखने पर, 1 = A    

∴ A = 1

t = -1 रखने पर, 1 = B(-1)  

∴ B = -1

`therefore 1/(t (t + 1)) = 1/t - 1/(t + 1)`

`therefore 1/2 int 1/(t (t + 1))  dt = 1/2 int 1/t dt - 1/2 int 1/(t + 1)  dt`

`= 1/2  log abs t - 1/2  log abs (t + 1) + C`

`= 1/2  log abs (x ^2) - 1/2  log abs(x ^2 + 1) + C`

`= log abs x - 1/2  log abs(x^2 + 1) + C`