integral of square root of (9u^4+4u^2+1)
To find the integral of √(9u^4 + 4u^2 + 1), we can use the method of substitution. Let's go step by step:
1. Start by making a substitution. Let's choose u^2 as our substitution variable.
Let u^2 = t.
Differentiating both sides with respect to u will give us du = (1/2t) dt.
2. Now, substitute these values in the given expression:
√(9u^4 + 4u^2 + 1) = √(9t^2 + 4t + 1) * (1/2t) dt.
3. Simplify the expression inside the square root:
√(9t^2 + 4t + 1) can be further simplified as √((3t + 1)^2).
4. Let's further simplify our integral:
∫ √((3t + 1)^2) * (1/2t) dt.
We simplified the integrand to (3t + 1)/2t.
5. Now, we can separate the integrand into two separate fractions:
∫ (3t/2t) + (1/2t) dt.
This can be split into two separate integrals:
∫ (3/2) dt + ∫ (1/2t) dt.
∫ (3/2) dt simplifies to (3/2)t + C₁ (integration with respect to t).
∫ (1/2t) dt simplifies to (1/2) ln |t| + C₂ (integration with respect to t).
C₁ and C₂ are constants of integration.
6. Substituting back u for t:
Finally, we substitute back u for our original variable:
(3/2)t + C₁ and (1/2) ln |t| + C₂ can be rewritten as:
(3/2)u^2 + C₁ and (1/2) ln |u^2| + C₂.
And that's the final result for the integral of √(9u^4 + 4u^2 + 1)!