How do I evaluate 2x∫1 3t(t^2 + 1)^2 dt?
I would expand out the integrand and then integrate each part.
Would my answer be -32x^6 - 48x^4 - 6x^2 + (7/2)then?
To evaluate the integral 2x∫1 3t(t^2 + 1)^2 dt, we can use the method of integration by substitution. Here's a step-by-step explanation of the process:
Step 1: Choose a substitute variable
Let's choose u = t^2 + 1 as our substitute variable. This choice will help simplify the integral.
Step 2: Compute du/dt
Differentiating both sides of the equation with respect to t, we get:
du/dt = 2t
Step 3: Rearrange to express dt in terms of du
Rearranging the differential equation, we get:
dt = du / (2t)
Step 4: Substitute the variables and convert the integral
Using the substitutions u = t^2 + 1 and dt = du / (2t), we can rewrite the integral as:
2x∫[1] 3t(t^2 + 1)^2 dt = 2x∫[1] 3t(u)^2 (du / (2t))
Now, we can simplify the expression:
2x∫[1] (3t^2)(u^2) (du / (2t))
Notice that the t in the numerator and denominator cancels out, simplifying the expression further:
2x∫[1] (3u^2) du
Step 5: Evaluate the integral
Integrating (3u^2) with respect to u, we get:
2x( u^3 )/3
Step 6: Substitute back in the original variable
Remembering that u = t^2 + 1, we substitute u back into the expression:
2x( (t^2 + 1)^3 )/3
So the integral evaluates to 2x( (t^2 + 1)^3 )/3.