10:[["$","title","0",{"children":"CUET Mathematics - e^x (1-x/1+x^2)^2 dx is equal to | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined The integral to solve is e^x (1-x/1+x^2)^2 dx For integrals with e^x, the formula e^x [f(x) + f'(x)] dx = e^x f(x) + C can be applied.

Expanding the square: (1-x/1+x^2)^2 = (1-x)^2/(1+x^2)^2 The integral becomes: e^x (1-x)^2/(1+x^2)^2 dx

Let f(x) = 1/1+x^2 Finding the derivative: f'(x) = -2x/(1+x^2)^2

Adding f(x) and f'(x): f(x) + f'(x) = 1/1+x^2 + -2x/(1+x^2)^2 = (1+x^2)/(1+x^2)^2 + -2x/(1+x^2)^2 = 1+x^2-2x/(1+x^2)^2 = 1-2x+x^2/(1+x^2)^2 = (1-x)^2/(1+x^2)^2 This matches the integrand.

Since (1-x)^2/(1+x^2)^2 = f(x) + f'(x) where f(x) = 1/1+x^2: e^x [(1-x)^2/(1+x^2)^2] dx = e^x f(x) + C = e^x/1+x^2 + C"}],["$","link","2",{"rel":"manifest","href":"https://www.afterboards.in/manifest.json","crossOrigin":"$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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - e^x (1-x/1+x^2)^2 dx is equal to | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined The integral to solve is e^x (1-x/1+x^2)^2 dx For integrals with e^x, the formula e^x [f(x) + f'(x)] dx = e^x f(x) + C can be applied.

Expanding the square: (1-x/1+x^2)^2 = (1-x)^2/(1+x^2)^2 The integral becomes: e^x (1-x)^2/(1+x^2)^2 dx

Let f(x) = 1/1+x^2 Finding the derivative: f'(x) = -2x/(1+x^2)^2

Adding f(x) and f'(x): f(x) + f'(x) = 1/1+x^2 + -2x/(1+x^2)^2 = (1+x^2)/(1+x^2)^2 + -2x/(1+x^2)^2 = 1+x^2-2x/(1+x^2)^2 = 1-2x+x^2/(1+x^2)^2 = (1-x)^2/(1+x^2)^2 This matches the integrand.

Since (1-x)^2/(1+x^2)^2 = f(x) + f'(x) where f(x) = 1/1+x^2: e^x [(1-x)^2/(1+x^2)^2] dx = e^x f(x) + C = e^x/1+x^2 + C"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-13/26"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - e^x (1-x/1+x^2)^2 dx is equal to | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined The integral to solve is e^x (1-x/1+x^2)^2 dx For integrals with e^x, the formula e^x [f(x) + f'(x)] dx = e^x f(x) + C can be applied.

Expanding the square: (1-x/1+x^2)^2 = (1-x)^2/(1+x^2)^2 The integral becomes: e^x (1-x)^2/(1+x^2)^2 dx

Let f(x) = 1/1+x^2 Finding the derivative: f'(x) = -2x/(1+x^2)^2

Adding f(x) and f'(x): f(x) + f'(x) = 1/1+x^2 + -2x/(1+x^2)^2 = (1+x^2)/(1+x^2)^2 + -2x/(1+x^2)^2 = 1+x^2-2x/(1+x^2)^2 = 1-2x+x^2/(1+x^2)^2 = (1-x)^2/(1+x^2)^2 This matches the integrand.

Since (1-x)^2/(1+x^2)^2 = f(x) + f'(x) where f(x) = 1/1+x^2: e^x [(1-x)^2/(1+x^2)^2] dx = e^x f(x) + C = e^x/1+x^2 + C"}],["$","meta","17",{"name":"twitter:image","content":"https://www.afterboards.in/pyp/AfterBoards%20CUET%20Past%20Year%20Papers%20with%20Solutions.webp"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f"]

CUET Mathematics 2025 26 May Shift 2 PYQs

Keyboard Shortcuts