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प्रश्न
Integrate the following w.r.t.x : cot–1 (1 – x + x2)
उत्तर
Let I = `int cot^-1 (1 - x + x^2)*dx`
= `int tan^-1 (1/(1 - x + x^2))*dx`
= `int tan^-1 [(x + (1 - x))/(1 - x(1 - x))]`
= `int [tan^-1 x + tan^-1 (1 - x)]*dx`
= `int tan^-1 x*dx + int tan^-1 (1 - x)*dx`
∴ I = I1 + I2 ...(1)
I1 = `int tan^-1 x*dx = int(tan^-1x)1*dx`
= `(tan^-1x)* int 1dx - [d/dx (tan^-1x)* int 1dx]*dx`
= `(tan^-1x)x - int 1/(1 + x^2)*x*dx`
= `xtan^-1 x - (1)/(2) int (2x)/(1 + x^2)*dx`
∴ I1 = `x tan^-1x - (1)/(2)log|1 + x^2| + c_1`
...`[because d/dx (1 + x^2) = 2x and int (f'(x))/f(x) dx = log|f(x)| + c]`
I2 = `int tan^-1 (1 - x)*dx`
= `int tan^-1 (1 - x)]*1dx`
= `[tan^-1 (1 - x)]*int 1dx - int {d/dx [tan^-1 (1 - x)]* int 1dx}*dx`
= `[tan^-1 (1 - x)]*x - int (1)/(1 + (1 - x)^2)*(-1)*xdx`
= `xtan^-1 (1 - x) + int x/(1 + 1 - 2x + x^2)*dx`
= `xtan^-1 (1 - x) + int x/(2 - 2x + x^2)*dx`
Let x = `"A"[d/dx (2 - 2x + x^2)] + "B"`
∴ x = A(– 2 + 2x) + B = 2Ax + (–2A + B)
Comparing the coefficient of x and constant on both the sides, we get
1 = 2A and 0 = – 2A + B
∴ A = `(1)/(2) and 0 = -2(1/2) + "B"`
∴ B = 1
∴ x = `(1)/(2)(- 2 + 2x) + 1`
∴ I2= `xtan^-1 (1 - x) + int (1/2(-2 + 2x) + 1)/(2 - 2x + x^2)*dx`
= `xtan^-1 (1 - x) + 1/2 (-2 + 2x)/(2 - 2x + x^2)*dx + int (1)/(2 - 2x + x^2)*dx`
= `xtan^-1 (1 - x) + (1)/(2) log|2 - 2x + x^2| + int (1)/(1 + (1 - 2x + x^2))*dx`
= `xtan^-1 (1 - x) + (1)/(2) log|x^2 - 2x + 2| + int (1)/(1 + (1 - x^2))*dx`
= `xtan^-1 (1 - x) + (1)/(2) log|x^2 - 2x + 2| + (1)/(1) (tan-1 (1 - x))/(-1) + c_2`
= `x tan^-1 (1 - x) + 1/2log|x^2 - 2x + 2| - tan^-1 (1 - x) + c_2`
= `(x - 1)tan^-1 (1 - x) + (1)/(2)log|x^2 - 2x + 2| + c_2`
∴ I2 = `-(1 - x)tan^-1 (1 - x) + (1)/(2)log|x^2 - 2x + 2| + c_2` ...(3)
From (1),(2) and (3), we get
I = `x tan^-1 x - (1)/(2) log|1 + x^2| + c_1 - (1 - x)tan^-1 (1 - x) + 1/2log|x^2 - 2x + 2| + c_2`
= `x tan^-1 x - (1)/(2) log|1 + x^2| - (1 - x)tan^-1 (1 - x) + 1/2 |x^2 - 2x + 2| + c`, where c = c1 + c2.
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