int(6x-7)/(4x^2-4x+5)dx=? 

দেওয়া আছে, ∫(6x-7)/(4x²-4x+5) dx

আমরা লিখতে পারি, 6x - 7 = A(d/dx(4x² - 4x + 5)) + B

=> 6x - 7 = A(8x - 4) + B

=> 6x - 7 = 8Ax - 4A + B

তুলনা করে পাই,

8A = 6 => A = 3/4

-4A + B = -7 => -4(3/4) + B = -7 => -3 + B = -7 => B = -4

সুতরাং, ∫(6x-7)/(4x²-4x+5) dx = ∫((3/4)(8x-4) - 4)/(4x²-4x+5) dx

= (3/4)∫(8x-4)/(4x²-4x+5) dx - 4∫1/(4x²-4x+5) dx

ধরি, I₁ = ∫(8x-4)/(4x²-4x+5) dx

4x² - 4x + 5 = u => (8x - 4)dx = du

=> I₁ = ∫1/u du = ln|u| + c₁ = ln|4x² - 4x + 5| + c₁

আবার, I₂ = ∫1/(4x²-4x+5) dx = ∫1/(4(x²-x+5/4)) dx

= ∫1/(4(x²-x+(1/2)²-(1/2)²+5/4)) dx = ∫1/(4((x-1/2)² + 1)) dx

= (1/4)∫1/((x-1/2)² + 1²) dx = (1/4)(1/1)tan^-1((x-1/2)/1) + c₂

= (1/4)tan^-1(x-1/2) + c₂

সুতরাং, ∫(6x-7)/(4x²-4x+5) dx = (3/4)ln|4x² - 4x + 5| - 4(1/4)tan^-1(x-1/2) + C

= (3/4)ln|4x² - 4x + 5| - tan^-1(x-1/2) + C

সুতরাং, নির্ণেয় সমাধান: (3/4)ln|4x² - 4x + 5| - tan^-1(x-1/2) + C 🥳🎉