Question

यदि `int ("d"x)/((x + 2) (x^2 + 1))= "a" log |1 + x^2| + "b" tan^-1x + 1/5 log |x + 2| + "C"` है, तो ______

विकल्प

• a = `(-1)/10`, b = `(-2)/5` 

• a = `1/10`, b = `- 2/5`

• a = `(-1)/10`, b = `2/5`

• a = `1/10`, b = `2/5`

Answer 1

यदि `int ("d"x)/((x + 2) (x^2 + 1))= "a" log |1 + x^2| + "b" tan^-1x + 1/5 log |x + 2| + "C"` है, तो `underline("a" = (-1)/10, "b" = 2/5)`

व्याख्या:

यह देखते हुए कि, `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`

अब, I = `int "dx"/((x + 2)(x^2 + 1))`

`1/((x + 2)(x^2 + 1)) = "A"/(x + 2) + ("B"x + "C")/(x^2 + 1)`

⇒ 1 = A(x² + 1) + (Bx + C)(x + 2)

⇒ 1 = (A + B)x² + (2B + C)x + A + 2C

गुणांक की तुलना करने पर, हम प्राप्त करते हैं

A + B = 0

A + 2C = 1

2B + C = 0

हल करने पर हमें A = `1/5` प्राप्त होता है,

B = `- 1/5`

और C = `2/5`

∴ `int "dx"/((x + 2)(x^2 + 1))`

= `1/5 int 1/(x + 2) "d"x + int (- 1/5 + 2/5)/(x^2 + 1) "d"x`

= `1/5 int 1/(x + 2) "d"x - 1/10 int (2x)/(1 + x^2) "d"x + 1/5 int 2/(1 + x^2) "d"x`

= `1/5 log|x + 2| - 1/10 log|1 + x^2| + 2/5 tan^-1x + "C"`

= `"a" log |1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`  ....(दिया गया)

∴ a = `(-1)/10`, b = `2/5`.