10:[["$","title","0",{"children":"CUET Mathematics - _0^1 tan^-1(2x-1/1+x-x^2)dx is equal to | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Looking at the expression inside: 2x-1/1+x-x^2 The formula for (A - B) is: (A - B) = A - B/1 + A B

Let A = x and B = 1-x The numerator becomes: A - B = x - (1-x) = x - 1 + x = 2x - 1 The denominator becomes: 1 + A B = 1 + x(1-x) = 1 + x - x^2

Since 2x-1/1+x-x^2 = A - B/1 + A B: ^-1(2x-1/1+x-x^2) = ^-1(x) - ^-1(1-x)

The integral becomes: _0^1 [^-1(x) - ^-1(1-x)]dx = _0^1 ^-1(x)dx - _0^1 ^-1(1-x)dx

For _0^1 ^-1(1-x)dx, <=t u = 1-x Then du = -dx When x = 0: u = 1 When x = 1: u = 0 _0^1 ^-1(1-x)dx = -_1^0 ^-1(u)du = _0^1 ^-1(u)du

_0^1 ^-1(x)dx - _0^1 ^-1(1-x)dx = _0^1 ^-1(x)dx - _0^1 ^-1(x)dx = 0 Therefore, the value of the integral is 0."}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$undefined"}],["$","meta","3",{"name":"keywords","content":"CUETMAT,MAT,Calculus,Integrals"}],["$","meta","4",{"name":"robots","content":"index, follow"}],["$","meta","5",{"name":"category","content":"education"}],["$","meta","6",{"name":"google-site-verification","content":"google"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - _0^1 tan^-1(2x-1/1+x-x^2)dx is equal to | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Looking at the expression inside: 2x-1/1+x-x^2 The formula for (A - B) is: (A - B) = A - B/1 + A B

Let A = x and B = 1-x The numerator becomes: A - B = x - (1-x) = x - 1 + x = 2x - 1 The denominator becomes: 1 + A B = 1 + x(1-x) = 1 + x - x^2

Since 2x-1/1+x-x^2 = A - B/1 + A B: ^-1(2x-1/1+x-x^2) = ^-1(x) - ^-1(1-x)

The integral becomes: _0^1 [^-1(x) - ^-1(1-x)]dx = _0^1 ^-1(x)dx - _0^1 ^-1(1-x)dx

For _0^1 ^-1(1-x)dx, <=t u = 1-x Then du = -dx When x = 0: u = 1 When x = 1: u = 0 _0^1 ^-1(1-x)dx = -_1^0 ^-1(u)du = _0^1 ^-1(u)du

_0^1 ^-1(x)dx - _0^1 ^-1(1-x)dx = _0^1 ^-1(x)dx - _0^1 ^-1(x)dx = 0 Therefore, the value of the integral is 0."}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-2/23"}],["$","meta","10",{"property":"og:site_name","content":"AfterBoards | CUET Mathematics Preparation"}],["$","meta","11",{"property":"og:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],["$","meta","12",{"property":"og:type","content":"website"}],["$","meta","13",{"name":"twitter:card","content":"summary_large_image"}],["$","meta","14",{"name":"twitter:site","content":"afterboards.in"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - _0^1 tan^-1(2x-1/1+x-x^2)dx is equal to | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Looking at the expression inside: 2x-1/1+x-x^2 The formula for (A - B) is: (A - B) = A - B/1 + A B

Let A = x and B = 1-x The numerator becomes: A - B = x - (1-x) = x - 1 + x = 2x - 1 The denominator becomes: 1 + A B = 1 + x(1-x) = 1 + x - x^2

Since 2x-1/1+x-x^2 = A - B/1 + A B: ^-1(2x-1/1+x-x^2) = ^-1(x) - ^-1(1-x)

The integral becomes: _0^1 [^-1(x) - ^-1(1-x)]dx = _0^1 ^-1(x)dx - _0^1 ^-1(1-x)dx

For _0^1 ^-1(1-x)dx, <=t u = 1-x Then du = -dx When x = 0: u = 1 When x = 1: u = 0 _0^1 ^-1(1-x)dx = -_1^0 ^-1(u)du = _0^1 ^-1(u)du

_0^1 ^-1(x)dx - _0^1 ^-1(1-x)dx = _0^1 ^-1(x)dx - _0^1 ^-1(x)dx = 0 Therefore, the value of the integral is 0."}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

CUET Mathematics 2025 13 May Shift 2 PYQs

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