integral of [root of(9t^4+4t^2+1)]

Question

please findout the integral value,

Answer 1

∫√(9t⁴+4t²+1)dt=3∫√((t²+2/9)²+5/81)dt.

(a²+b)ⁿ=(b+a²)ⁿ=bⁿ(1+a²/b)ⁿ=

bⁿ(1+na²/b+(n(n-1)/2!)(a⁴/b²)+(n(n-1)(n-2)/3!)(a⁶/b³)+...) (binomial expansion)

When n=½, the coefficients become ½, -⅛, 1/16, -35/128, etc.

(a²+b)^½=(b+a²)^½=√b(1+a²/b)^½=

√b(1+a²/(2b)-a⁴/(8b²)+a⁶/(16b³)-35a⁸/(128b⁴)...) (binomial expansion)

When a=t²+2/9 and b=5/81, the powers of a are powers of (t²+2/9) and can be integrated, while the powers of b are numerical and can be calculated.

For example, (t²+2/9)²=t⁴+4t²/9+4/81 can be integrated: t⁵/5+4t³/27+4t/81, and 1/(2b)=81/10; √b=√5/9. Thus, the series can be evaluated, even though the process is cumbersome. Example: a²/(2b)=81t⁵/50+6t³/5+2t/5.

So the integral can be resolved through a power series, although with difficulty.

A different approach may be obtained by approximation:

9t⁴+4t²+1=(3t²+⅔)²+5/9, so √(9t⁴+4t²+1)=(3t²+⅔)(1+5/(9(3t²+⅔)²)^½.

An approximation to this is: (3t²+⅔)(1+5/(18(3t²+⅔)²)=3t²+⅔+5/(18(3t²+⅔)) which "integrates":

t³+2t/3+(5/18)∫dt/(3t²+⅔). If t is small this becomes t³+2t/3+5/12, and if t is large it becomes t³+2t/3-5/(54t).

So the integral is between t³+2t/3-5/(54t) and t³+2t/3+5/12.

In the meantime, I'm on the lookout for a more precise method...