10:[["$","title","0",{"children":"CUET Mathematics - sqrt(1 + x^2/9) dx is equal to (Where C is an arbitrary constant) | PYQs + Solutions | AfterBoards"}],["$","meta","1",{"name":"description","content":"undefined Simplify the square root: sqrt(1 + x^2/9) = sqrt(9 + x^2/9) = sqrt(9 + x^2)sqrt(9) = sqrt(x^2 + 9)3 = 1/3sqrt(x^2 + 9)

The integral becomes: sqrt(1 + x^2/9) dx = 1/3sqrt(x^2 + 9) dx = 1/3 sqrt(x^2 + 9) dx

This matches the standard form sqrt(x^2 + a^2) dx where a^2 = 9, so a = 3. The standard formula is: sqrt(x^2 + a^2) dx = x/2sqrt(x^2 + a^2) + a^2/2|x + sqrt(x^2 + a^2)| + C

Substituting a = 3: sqrt(x^2 + 9) dx = x/2sqrt(x^2 + 9) + 9/2|x + sqrt(x^2 + 9)| + C

Multiplying by 1/3: sqrt(1 + x^2/9) dx = 1/3[x/2sqrt(x^2 + 9) + 9/2|x + sqrt(x^2 + 9)|] + C = 1/3[9/2|x + sqrt(x^2 + 9)| + x/2sqrt(x^2 + 9)] + C Therefore, the answer is: 1/3[9/2|x + sqrt(x^2 + 9)| + x/2sqrt(x^2 + 9)] + 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"}],["$","meta","7",{"property":"og:title","content":"CUET Mathematics - sqrt(1 + x^2/9) dx is equal to (Where C is an arbitrary constant) | PYQs + Solutions"}],["$","meta","8",{"property":"og:description","content":"undefined Simplify the square root: sqrt(1 + x^2/9) = sqrt(9 + x^2/9) = sqrt(9 + x^2)sqrt(9) = sqrt(x^2 + 9)3 = 1/3sqrt(x^2 + 9)

The integral becomes: sqrt(1 + x^2/9) dx = 1/3sqrt(x^2 + 9) dx = 1/3 sqrt(x^2 + 9) dx

This matches the standard form sqrt(x^2 + a^2) dx where a^2 = 9, so a = 3. The standard formula is: sqrt(x^2 + a^2) dx = x/2sqrt(x^2 + a^2) + a^2/2|x + sqrt(x^2 + a^2)| + C

Substituting a = 3: sqrt(x^2 + 9) dx = x/2sqrt(x^2 + 9) + 9/2|x + sqrt(x^2 + 9)| + C

Multiplying by 1/3: sqrt(1 + x^2/9) dx = 1/3[x/2sqrt(x^2 + 9) + 9/2|x + sqrt(x^2 + 9)|] + C = 1/3[9/2|x + sqrt(x^2 + 9)| + x/2sqrt(x^2 + 9)] + C Therefore, the answer is: 1/3[9/2|x + sqrt(x^2 + 9)| + x/2sqrt(x^2 + 9)] + C"}],["$","meta","9",{"property":"og:url","content":"https://www.afterboards.in/past-year-questions/cuet-mathematics/2025/slot-19/5"}],["$","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"}],["$","meta","15",{"name":"twitter:title","content":"CUET Mathematics - sqrt(1 + x^2/9) dx is equal to (Where C is an arbitrary constant) | PYQs + Solutions"}],["$","meta","16",{"name":"twitter:description","content":"undefined Simplify the square root: sqrt(1 + x^2/9) = sqrt(9 + x^2/9) = sqrt(9 + x^2)sqrt(9) = sqrt(x^2 + 9)3 = 1/3sqrt(x^2 + 9)

The integral becomes: sqrt(1 + x^2/9) dx = 1/3sqrt(x^2 + 9) dx = 1/3 sqrt(x^2 + 9) dx

This matches the standard form sqrt(x^2 + a^2) dx where a^2 = 9, so a = 3. The standard formula is: sqrt(x^2 + a^2) dx = x/2sqrt(x^2 + a^2) + a^2/2|x + sqrt(x^2 + a^2)| + C

Substituting a = 3: sqrt(x^2 + 9) dx = x/2sqrt(x^2 + 9) + 9/2|x + sqrt(x^2 + 9)| + C

CUET Mathematics 2025 3 June Shift 1 PYQs

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