integrate ((x^3/6)+(1/(2x)))*squareroot of 1+(x^4/4)+(1/4x^2)-1/2..plz help me...tnx in advance!
To integrate the given function, we can use the technique of substitution. Let's break down the steps:
Step 1: Simplify the expression
Start by simplifying the expression under the square root:
1 + (x^4/4) + (1/4x^2) - 1/2
= (4x^2 + x^4 + 1 - 2x^2) / 4x^2
= (x^4 + 2x^2 + 1) / 4x^2
Step 2: Rewrite the expression using fractional exponents
Rewrite the expression (x^4 + 2x^2 + 1) / 4x^2 using fractional exponents:
= (x^4 + 2x^2 + 1) / (4x^2)^(1/2)
= (x^4 + 2x^2 + 1) / (2x)^(2*(1/2))
= (x^4 + 2x^2 + 1) / (2x)^(2/2)
= (x^4 + 2x^2 + 1) / (2x)
Step 3: Apply substitution
Let's substitute u = x^2:
= (u^2 + 2u + 1) / (2x)
Then differentiate both sides with respect to x:
du/dx = 2x
Now we can rewrite the integral in terms of u:
∫ ((x^3/6) + (1/(2x))) * √(1 + (x^4/4) + (1/4x^2) - 1/2) dx
= ∫ ((u/6) + (1/(4u))) * √(u + 2) * (1/2x) du
= (1/12) ∫ ((u/√(u+2)) + (1/(2u√(u+2)))) du
Step 4: Integrate the simplifed expression
Now we can integrate the simplified expression:
(1/12) ∫ ((u/√(u+2)) + (1/(2u√(u+2)))) du
The integral of u/√(u+2) can be found using the substitution v = u + 2:
= (1/12) ∫ ((v - 2)/√v) dv
= (1/12) ∫ (v^(1/2) - 2v^(-1/2)) dv
= (1/12) * (2/3 * v^(3/2) - 4 * v^(1/2)) + C
= (1/18) * v^(3/2) - (1/3) * v^(1/2) + C
Using the substitution v = u + 2 again, substitute back in u:
= (1/18) * (u + 2)^(3/2) - (1/3) * (u + 2)^(1/2) + C
Therefore, the final result is:
∫ ((x^3/6) + (1/(2x))) * √(1 + (x^4/4) + (1/4x^2) - 1/2) dx
= (1/18) * (x^2 + 2)^(3/2) - (1/3) * (x^2 + 2)^(1/2) + C
Please note that C represents the constant of integration, which is added to indefinite integrals.