10:[["$","title","0",{"children":"CUET Mathematics - dxx^3sqrt((1 + x^4)) = | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Find dxx^3sqrt((1 + x^4))

Looking at the integral with x^3 in the denominator and x^4 inside the square root, use the substitution t = 1/x^2.

If t = 1/x^2 = x^-2 dt/dx = -2x^-3 dt/dx = -2/x^3 dt = -2/x^3dx dx/x^3 = -1/2dt

Since t = 1/x^2, then x^2 = 1/t and x^4 = 1/t^2 sqrt(1 + x^4) = sqrt(1 + 1/t^2) = sqrt(t^2 + 1/t^2) = sqrt(t^2 + 1)t (assuming x > 0, so t > 0)

The integral becomes: 1sqrt(t^2 + 1)t (-1/2)dt = -1/2 tsqrt(t^2 + 1) dt

Let v = t^2 + 1 dv = 2t dt t dt = 1/2dv The integral becomes: -1/2 1sqrt(v) 1/2dv = -1/4 v^-1/2dv

-1/4 v^1/21/2 + C = -1/4 2sqrt(v) + C = -1/2sqrt(v) + C

Replace v = t^2 + 1: -1/2sqrt(t^2 + 1) + C Replace t = 1/x^2: -1/2sqrt(1/x^4 + 1) + C = -1/2sqrt(1 + x^4/x^4) + C = -1/2 sqrt(1 + x^4)x^2 + C = -sqrt(1 + x^4)2x^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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - dxx^3sqrt((1 + x^4)) = | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Find dxx^3sqrt((1 + x^4))

Looking at the integral with x^3 in the denominator and x^4 inside the square root, use the substitution t = 1/x^2.

If t = 1/x^2 = x^-2 dt/dx = -2x^-3 dt/dx = -2/x^3 dt = -2/x^3dx dx/x^3 = -1/2dt

Since t = 1/x^2, then x^2 = 1/t and x^4 = 1/t^2 sqrt(1 + x^4) = sqrt(1 + 1/t^2) = sqrt(t^2 + 1/t^2) = sqrt(t^2 + 1)t (assuming x > 0, so t > 0)

The integral becomes: 1sqrt(t^2 + 1)t (-1/2)dt = -1/2 tsqrt(t^2 + 1) dt

Let v = t^2 + 1 dv = 2t dt t dt = 1/2dv The integral becomes: -1/2 1sqrt(v) 1/2dv = -1/4 v^-1/2dv

-1/4 v^1/21/2 + C = -1/4 2sqrt(v) + C = -1/2sqrt(v) + C

Replace v = t^2 + 1: -1/2sqrt(t^2 + 1) + C Replace t = 1/x^2: -1/2sqrt(1/x^4 + 1) + C = -1/2sqrt(1 + x^4/x^4) + C = -1/2 sqrt(1 + x^4)x^2 + C = -sqrt(1 + x^4)2x^2 + C"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-12/21"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - dxx^3sqrt((1 + x^4)) = | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Find dxx^3sqrt((1 + x^4))

Looking at the integral with x^3 in the denominator and x^4 inside the square root, use the substitution t = 1/x^2.

If t = 1/x^2 = x^-2 dt/dx = -2x^-3 dt/dx = -2/x^3 dt = -2/x^3dx dx/x^3 = -1/2dt

Since t = 1/x^2, then x^2 = 1/t and x^4 = 1/t^2 sqrt(1 + x^4) = sqrt(1 + 1/t^2) = sqrt(t^2 + 1/t^2) = sqrt(t^2 + 1)t (assuming x > 0, so t > 0)

The integral becomes: 1sqrt(t^2 + 1)t (-1/2)dt = -1/2 tsqrt(t^2 + 1) dt

Let v = t^2 + 1 dv = 2t dt t dt = 1/2dv The integral becomes: -1/2 1sqrt(v) 1/2dv = -1/4 v^-1/2dv

-1/4 v^1/21/2 + C = -1/4 2sqrt(v) + C = -1/2sqrt(v) + C

Replace v = t^2 + 1: -1/2sqrt(t^2 + 1) + C Replace t = 1/x^2: -1/2sqrt(1/x^4 + 1) + C = -1/2sqrt(1 + x^4/x^4) + C = -1/2 sqrt(1 + x^4)x^2 + C = -sqrt(1 + x^4)2x^2 + C"}],"$L17","$L18","$L19","$L1a","$L1b","$L1c","$L1d","$L1e","$L1f","$L20"]

CUET Mathematics 2025 22 May Shift 2 PYQs

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