Integral Request

Could someone with a CAS please evaluate and copy the integration steps (I already know the answer) to the integral from 0 to 1 S(2t+9t^3)sqrt(1+4t^2+9t^4)dt please?

It sure would be nice if this site supported LaTeX.

I think you need to use substitution where you equate the derivative of the group that has the largest exponent to the other one.

Deriv of  (1+4t^2+9t^4) is (8t + 36t^3)

Compare that to the other thing which is (2t + 9t^3)

Notice how the deriv of the 1st one is 4 times greater than the 2t + 9t^3

Thus integrate 1/4(1+4t^2+9t^4) = 1/2 (1+4t^2+9t^4)^2

Then from 0 to 1 gives 97.5 I think...

I'm not sure if that's right or not... I'm not a Calculus expert but know a little... hopefully this is accurate. If not, what is the answer?

EIther way I can use Maple tomorrow, but Wolfram Alpha gave this answer without steps:

http://www.wolframalpha.com/input/?i=integrate+from+0+to+1+%282t%2B9t^3%29sqrt%281%2B4t^2%2B9t^4%29dt

I guess it's too late, but I'll write it anyway:

Integral(0->1; (2t+9t^3)SQRT(1+4t^2+9t^4);  dt)= [x=1+4t^2+9t^4; dt=dx/4(2t+9t^3); t=0 => x=1; t=1 => x=14] = Integral(1->14; (2t+9t^3)SQRT(x)/4(2t+9t^3);  dx)= =0.25*Integral(1->14; SQRT(x);  dx)= 0.25 * 2/3 * 14SQRT(14) - 0.25 * 2/3 * 1SQRT(1)= =7SQRT(14)/3 - 1/6, As Wolfram Alpha suggests.

Hope anyone can understand anything when it's written like that, though.